Mechanical Waves

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This document introduces the concept of mechanical waves, a fundamental area of study in physics with far-reaching implications. Mechanical waves, such as those observed in earthquakes or musical instruments, require a medium to propagate, distinguishing them from electromagnetic waves. This work builds upon the principles of mechanics and oscillations, extending these concepts to wave phenomena. Prior research in areas like simple harmonic motion and the properties of materials provides a foundation for understanding how mechanical waves travel and interact. The study of mechanical waves is crucial for understanding diverse phenomena, from the transmission of seismic waves during earthquakes to the production of sound in musical instruments, and even the design of structures to withstand vibrations.

Image summary: The image is a photograph. It depicts significant destruction in an urban environment. The photo shows a collapsed building with debris scattered across the foreground, and other buildings are visible in the background. The extensive damage suggests a catastrophic event, such as a powerful earthquake or explosion, has occurred.

Mechanical Waves

Learning Outcomes

In this chapter, you'll learn...

15.1 What is meant by a mechanical wave, and the different varieties of mechanical waves.

15.2 How to use the relationship among speed, frequency, and wavelength for a periodic wave.

15.3 How to interpret and use the mathematical expression for a sinusoidal periodic wave.

15.4 How to calculate the speed of waves on a rope or string.

15.5 How to calculate the rate at which a mechanical wave transports energy.

15.6 What happens when mechanical waves overlap and interfere.

15.7 The properties of standing waves on a string, and how to analyze these waves.

15.8 How stringed instruments produce sounds of specific frequencies.

You'll need to review...

8.1 The impulse-momentum theorem.

14.1, 14.2 Periodic motion and simple harmonic motion. ripples on a pond, musical sounds, seismic tremors triggered by an earthquake—all these are wave phenomena. Waves can occur whenever a system is disturbed from equilibrium and when the disturbance can travel, or propagate, from one region of the system to another. As a wave propagates, it carries energy. The energy in light waves from the sun warms the surface of our planet; the energy in seismic waves can crack our planet's crust.

2 Definitions

Definition 1: Medium: A material or substance through which a wave propagates.

Definition 2: Sinusoidal Wave: A periodic wave with a shape that can be described by a sine or cosine function, exhibiting simple harmonic motion.

This chapter and the next are about mechanical waves—waves that travel within some material called a medium. (Chapter 16 is concerned with sound, an important type of mechanical wave.) We'll begin this chapter by deriving the basic equations for describing waves, including the important special case of sinusoidal waves in which the wave pattern is a repeating sine or cosine function. To help us understand waves in general, we'll look at the simple case of waves that travel on a stretched string or rope.

Waves on a string are important in music. When a musician strums a guitar or bows a violin, she makes waves that travel in opposite directions along the instrument's strings. What happens when these oppositely directed waves overlap is called interference. We'll discover that sinusoidal waves can occur on a guitar or violin string only for certain special frequencies, called normal-mode frequencies, determined by the properties of the string. These normal-mode frequencies determine the pitch of the musical sounds that a stringed instrument produces. (In the next chapter we'll find that interference also helps explain the pitches of wind instruments such as pipe organs.)

Not all waves are mechanical in nature. Electromagnetic waves—including light, radio waves, infrared and ultraviolet radiation, and x rays—can propagate even in empty space, where there is no medium. We'll explore these and other nonmechanical waves in later chapters.

15.1 Types of Mechanical Waves

A mechanical wave is a disturbance that travels through some material or substance called the medium for the wave. As the wave travels through the medium, the particles that make up the medium undergo displacements of various kinds, depending on the nature of the wave.

Figure 15.1 shows three varieties of mechanical waves. In figure 15.1a the medium is a string or rope under tension. If we give the left end a small upward shake or wiggle, the wiggle travels along the length of the string.

Figure 15.1 summary: The figure includes multiple illustrations depicting different types of waves. The first illustration shows a transverse wave on a string, where the particles of the string move perpendicularly to the direction of wave motion. Another illustration shows a longitudinal wave in a fluid, where the particles of the fluid move parallel to the direction of wave motion. The final illustration shows surface particles of a liquid moving in a circle as the wave passes. The figure demonstrates the difference in particle motion between transverse and longitudinal waves, as well as the circular motion of particles in a surface wave.

Definition

Transverse Wave: A wave in which the displacement of the medium is perpendicular to the direction of propagation.

Successive sections of string go through the same motion that we gave to the end, but at successively later times. Because the displacements of the medium are perpendicular or transverse to the direction of travel of the wave along the medium, this is called a transverse wave.

Definition

Longitudinal Wave: A wave in which the displacement of the medium is parallel to the direction of propagation.

In figure 15.1b the medium is a liquid or gas in a tube with a rigid wall at the right end and a movable piston at the left end. If we give the piston a single back-and-forth motion, displacement and pressure fluctuations travel down the length of the medium. This time the motions of the particles of the medium are back and forth along the same direction that the wave travels. We call this a longitudinal wave.

In figure 15.1c the medium is a liquid in a channel, such as water in an irrigation ditch or canal. When we move the flat board at the left end forward and back once, a wave disturbance travels down the length of the channel. In this case the displacements of the water have both longitudinal and transverse components.

Each of these systems has an equilibrium state. For the stretched string it is the state in which the system is at rest, stretched out along a straight line. For the fluid in a tube it is a state in which the fluid is at rest with uniform pressure.

And for the liquid in a trough it is a smooth, level water surface. In each case the wave motion is a disturbance from equilibrium that travels from one region of the medium to another. And in each case forces tend to restore the system to its equilibrium position when it is displaced, just as the force of gravity tends to pull a pendulum toward its straight-down equilibrium position when it is displaced.

These examples have three things in common. First, in each case the disturbance travels or propagates with a definite speed through the medium. This speed is called the speed of propagation, or simply the wave speed. Its value is determined in each case by the mechanical properties of the medium.

We'll use the symbol v for wave speed. (The wave speed is not the same as the speed with which particles move when they are disturbed by the wave. We'll return to this point in Section 15.3.) Second, the medium itself does not travel through space; its individual particles undergo back-and-forth or up-and-down motions around their equilibrium positions. The overall pattern of the wave disturbance is what travels.

Third, to set any of these systems into motion, we have to put in bio application Waves on a Snake's Body A snake moves itself along the ground by producing waves that travel backward along its body from its head to its tail. The waves remain stationary with respect to the ground as they push against the ground, so the snake moves forward. energy by doing mechanical work on the system. The wave motion transports this energy from one region of the medium to another. Waves transport energy, but not matter, from one region to another (Fig. 15.2).

Image summary:This is a photograph. The image shows a snake on a rock. The snake appears to be blending with the environment. The snake is likely in its natural habitat.

Figure 15.2 summary: The figure is a photograph. The photograph depicts a large crowd of people at a sports stadium performing "the wave". In "the wave", the disturbance propagates through the crowd, but the people stay in their seats. This illustrates a mechanical wave, where energy is transferred without the transport of matter.

test your understanding of section 15.1 What type of wave is “the wave” shown in figure 15.2? (1) Transverse; (2) longitudinal; (3) a combination of transverse and longitudinal.

(1) The “wave” travels horizontally from one spectator to the next along each row of the stadium, but the displacement of each spectator is vertically upward. Since the displacement is perpendicular to the direction in which the wave travels, the wave is transverse.

15.2 Periodic Waves

Definition

Wave Pulse: A single, non-repeating disturbance that travels through a medium.

The transverse wave on a stretched string in figure 15.1a is an example of a wave pulse. The hand exerts a transverse force that shakes the string up and down just once, producing a single “wiggle,” or pulse, that travels along the length of the string. The tension in the string restores its straight-line shape once the pulse has passed.

Definition

Periodic Wave: A wave in which the disturbance repeats at regular intervals.

A more interesting situation develops when we give the free end of the string a repetitive, or periodic, motion. (You should review the discussion of periodic motion in Chapter 14 before going ahead.) Each particle in the string undergoes periodic motion as the wave propagates, and we have a periodic wave.

Periodic Transverse Waves

Definition

Harmonic: An integer multiple of the fundamental frequency in a periodic wave.

Suppose we move one end of the string up and down with simple harmonic motion (S.H.M.) as in figure 15.3, with amplitude A, frequency f, angular frequency omega = 2pi f , and period T = 1/f = 2pi/omega . The wave that results is a symmetric sequence of crests and troughs. As we'll see, periodic waves with S.H.M. are particularly easy to analyze; we call them sinusoidal waves. It turns out that any periodic wave can be represented as a combination of sinusoidal waves. So this kind of wave motion is worth special attention.

Figure 15.3 summary: The figure is an illustration of a block-spring system undergoing simple harmonic motion and generating a sinusoidal wave on a string. The illustration depicts the motion of the wave, highlighting the crests, troughs, and amplitude. The simple harmonic motion of the spring and mass system generates a sinusoidal wave in the string, where each particle in the string exhibits the same harmonic motion as the spring and mass. The amplitude of the wave corresponds to the amplitude of this motion.

In figure 15.3 the wave is a continuous succession of transverse sinusoidal disturbances. Figure 15.4 shows the shape of a part of the string near the left end at time intervals of 1 over 8 of a period, for a total time of one period. The wave shape advances steadily toward the right, as indicated by the highlighted area.

Figure 15.4 summary: The figure is a depiction of a sinusoidal transverse wave as it travels to the right along a string. The figure displays the wave at different time instances, starting from time zero to a full period T. The wave is shown to progress along the x-axis over time. The wave advances by one wavelength during each period. The wave's displacement along the y-axis varies sinusoidally with both position and time.

As the wave moves, any point on the string (any of the red dots, for example) oscillates up and down about its equilibrium position with simple harmonic motion. When a sinusoidal wave passes through a medium, every particle in the medium undergoes simple harmonic motion with the same frequency.

caution Wave motion versus particle motion Don't confuse the motion of the transverse wave along the string and the motion of a particle of the string. The wave moves with constant speed v along the length of the string, while the motion of the particle is simple harmonic and transverse (perpendicular) to the length of the string.

For a periodic wave, the shape of the string at any instant is a repeating pattern. The wavelength lambda (the Greek letter lambda) of the wave is the distance from one crest to the next, or from one trough to the next, or from any point to the corresponding point on the next repetition of the wave shape. The wave pattern travels with constant speed v and advances a distance of one wavelength lambda in a time interval of one period T. So the wave speed is v = lambda/T or, because f = 1/T from equation (14.1),

Math summary: This equation calculates the wave speed of a periodic wave. It multiplies the wavelength, which is the distance between repetitions of the wave, by the frequency, which is how often the wave repeats.

The speed of propagation equals the product of wavelength and frequency. The frequency is a property of the entire periodic wave because all points on the string oscillate with the same frequency f.

Waves on a string propagate in just one dimension (in figure 15.4, along the x-axis). But the ideas of frequency, wavelength, and amplitude apply equally well to waves that propagate in two or three dimensions. Figure 15.5 shows a wave propagating in two dimensions on the surface of a tank of water. As with waves on a string, the wavelength is the distance from one crest to the next, and the amplitude is the height of a crest above the equilibrium level.

Figure 15.5 summary: This is an image that illustrates wave propagation in water. The image depicts circular waves emanating from a central point where a droplet impacts the water surface. The figure identifies the crests and troughs of the waves. The figure also indicates the wavelength.

In many important situations including waves on a string, the wave speed v is determined entirely by the mechanical properties of the medium. In this case, increasing f causes lambda to decrease so the product v = lambda f remains the same, and waves of all frequencies propagate with the same wave speed. In this chapter we'll consider only waves of this kind. (In later chapters we'll study the propagation of light waves in transparent materials where the wave speed depends on frequency; this turns out to be the reason raindrops create a rainbow.)

Figure 15.6 Using an oscillating piston to make a sinusoidal longitudinal wave in a fluid.

2 Definitions

Definition 1: Compression: A region of increased density and pressure in a longitudinal wave.

Definition 2: Rarefaction: A region of decreased density and pressure in a longitudinal wave.

Forward motion of the plunger creates a compression (a zone of high density); backward motion creates a rarefaction (a zone of low density).

Periodic Longitudinal Waves

To understand the mechanics of a periodic longitudinal wave, consider a long tube filled with a fluid, with a piston at the left end as in figure 15.1b. If we push the piston in, we compress the fluid near the piston, increasing the pressure in this region. This region then pushes against the neighboring region of fluid, and so on, and a wave pulse moves along the tube.

Now suppose we move the piston back and forth in S.H.M. along a line parallel to the axis of the tube (Fig. 15.6). This motion forms regions in the fluid where the pressure and density are greater or less than the equilibrium values. We call a region of increased density a compression; a region of reduced density is a rarefaction. Figure 15.6 shows compressions as darkly shaded areas and rarefactions as lightly shaded areas. The wavelength is the distance from one compression to the next or from one rarefaction to the next.

Figure 15.6 summary: The figure is a diagram that illustrates a sinusoidal longitudinal wave as it travels through a fluid. The diagram depicts a piston oscillating in simple harmonic motion, creating areas of compression and rarefaction in the fluid. The compression regions represent areas of higher density and pressure, while the rarefaction regions represent areas of lower density and pressure. The wave travels to the right with a certain speed. The wave has the same amplitude and period as the oscillation of the piston. The wavelength is the distance between two consecutive compressions or rarefactions.

Definition

Phase: The argument of the cosine or sine function in a wave function, determining the state of oscillation at a particular point and time.

Figure 15.7 shows the wave propagating in the fluid-filled tube at time intervals of 1 over 8 of a period, for a total time of one period. The pattern of compressions and rarefactions moves steadily to the right, just like the pattern of crests and troughs in a sinusoidal transverse wave (compare figure 15.4). Each particle in the fluid oscillates in S.H.M. parallel to the direction of wave propagation (that is, left and right) with the same amplitude A and period T as the piston. The particles shown by the two red dots in figure 15.7 are one wavelength apart, and so oscillate in phase with each other.

Figure 15.7 summary: The figure illustrates a sinusoidal longitudinal wave propagating through a fluid, created by a piston oscillating in simple harmonic motion. The graphic depicts the wave's progression over one full period, showing the positions of fluid particles at different time intervals. It can be inferred that the wave's amplitude and period are directly related to the piston's oscillation, and the particles oscillate around their equilibrium positions as the wave passes, demonstrating the wave's longitudinal nature.

Just like the sinusoidal transverse wave shown in figure 15.4, in one period T the longitudinal wave in figure 15.7 travels one wavelength lambda to the right. Hence the fundamental equation v = lambda f holds for longitudinal waves as well as for transverse waves, and indeed for all types of periodic waves. Just as for transverse waves, in this chapter and the next we'll consider only situations in which the speed of longitudinal waves does not depend on the frequency.

Example 15.1 Wavelength of a Musical Sound

With Variation Problems

Sound waves are longitudinal waves in air. The speed of sound depends on temperature; at 20 sup circle C it is 344 meters per second (1130 feet per second). What is the wavelength of a sound wave in air at 20 sup circle C if the frequency is 262 Hz (the approximate frequency of middle C on a piano)?

identify and set up This problem involves equation (15.1), v equals lambda times f, which relates wave speed v, wavelength lambda, and frequency f for a periodic wave. The target variable is the wavelength lambda. We are given v equals 344 meters per second and f equals 262 Hertz equals 262 times s to the power of negative 1.

Math summary: This expression calculates the wavelength of a wave by dividing its speed by its frequency. The wave speed of 344 meters per second is divided by a frequency of 262 hertz, resulting in a wavelength of 1.31 meters.

evaluate The speed v of sound waves does not depend on the frequency. Hence lambda equals v divided by f says that wavelength changes in inverse proportion to frequency. As an example, high (soprano) C is two octaves above middle C. Each octave corresponds to a factor of 2 in frequency, so the frequency of high C is four times that of middle C: f equals 4 times 262 Hertz equals 1048 Hertz. Hence the wavelength of high C is one-fourth as large: lambda equals 1.31 meters divided by 4 equals 0.328 meters.

execute We solve Equation fifteen point one for lambda:

keyconcept The product of a wave's wavelength and frequency has the same value no matter what the frequency is. This product equals the wave speed.

test your understanding of section 15.2 If you double the wavelength of a wave on a particular string without changing the physical properties of the string, what happens to the wave speed v and the frequency f? (1) v doubles and f is unchanged; (2) v is unchanged and f doubles; (3) v becomes one-half as great and f is unchanged; (4) v is unchanged and f becomes one-half as great; (v) none of these.

Answer

(4) The speed of waves on a string, v, does not depend on the wavelength. We can rewrite the relationship v equals lambda times f as f equals v divided by lambda, which tells us that if the wavelength lambda doubles, the frequency f

15.3 Mathematical Description of a Wave

Many characteristics of periodic waves can be described by using the concepts of wave speed, amplitude, period, frequency, and wavelength. Often, though, we need a more detailed description of the positions and motions of individual particles of the medium at particular times during wave propagation.

As a specific example, let's look at waves on a stretched string. If we ignore the sag of the string due to gravity, the equilibrium position of the string is along a straight line. We take this to be the x-axis of a coordinate system.

Waves on a string are transverse; during wave motion a particle with equilibrium position x is displaced some distance y in the direction perpendicular to the x-axis. The value of y depends on which particle we are talking about (that is, y depends on 10) and also on the time t when we look at it. Thus y is a function of both x and t; y = y(x, t) . We call y(x, t) the wave function that describes the wave. If we know this function for a particular wave motion, we can use it to find the displacement (from equilibrium) of any particle at any time.

From this we can find the velocity and acceleration of any particle, the shape of the string, and anything else we want to know about the behavior of the string at any time.

Wave Function for a Sinusoidal Wave

Let's see how to determine the form of the wave function for a sinusoidal wave. Suppose a sinusoidal wave travels from left to right (the direction of increasing 10) along the string, as in figure 15.8. Every particle of the string oscillates in simple harmonic motion with the same amplitude and frequency. But the oscillations of particles at different points on the string are not all in step with each other.

Figure 15.8 summary: The figure is a depiction of a sinusoidal wave propagating along a string, shown at different time intervals. The wave's progression is illustrated by tracking the oscillations of three distinct points on the string. As time advances, the wave moves forward, causing each of the marked points to oscillate vertically. The movement of the points demonstrates the wave's periodic nature and its progression along the string, illustrating how a wave travels through a medium by the oscillation of its particles.

The particle at point B in figure 15.8 is at its maximum positive value of y at t = 0 and returns to y = 0 at t = 7 over 8T ; these same events occur for a particle at point A or point C at t = 4 over 8T and t = 6 over 8T , exactly one half-period later. For any two particles of the string, the motion of the particle on the right (in terms of the wave, the “downstream” particle) lags behind the motion of the particle on the left by an amount proportional to the distance between the particles.

Definition

Phase Difference: The difference in the phase of motion between two particles in a wave.

Hence the cyclic motions of various points on the string are out of step with each other by various fractions of a cycle. We call these differences phase differences, and we say that the phase of the motion is different for different points. For example, if one point has its maximum positive displacement at the same time that another has its maximum negative displacement, the two are a half-cycle out of phase. (This is the case for points A and B, or points B and C.)

Suppose that the displacement of a particle at the left end of the string (x = 0) , where the wave originates, is given by

Math summary: This expression calculates the displacement of a particle at a specific location and time. It takes the amplitude, angular frequency, and time as inputs to compute the displacement using a cosine function.

That is, the particle oscillates in S.H.M. with amplitude A, frequency f, and angular frequency omega = 2pi f . The notation y(x = 0, t) reminds us that the motion of this particle is a special case of the wave function y(x, t) that describes the entire wave. At t = 0 the particle at x = 0 is at its maximum positive displacement (y = A) and is instantaneously at rest (because y is a maximum).

The wave disturbance travels from x = 0 to some point 10 to the right of the origin in an amount of time given by x/v, where v is the wave speed. So the motion of point 10 at time t is the same as the motion of point 10 equals 0 at the earlier time t minus x divided by v. Hence we can find the displacement of point 10 at time t by simply replacing t in Equation fifteen point two by the quantity t minus x divided by v.

Figure 15.9 Two graphs of the wave function y(x, t) in equation (15.7). (a) Graph of displacement y versus coordinate x at time t = 0. (b) Graph of displacement y versus time t at coordinate x = 0. The vertical scale is exaggerated in both (a) and (b).

Figure 15.9 summary: The figure displays two sinusoidal wave plots. The first plot illustrates the wave's displacement with respect to position, showing the wavelength. The second plot shows the wave's displacement with respect to time, indicating the period. The wave exhibits a repetitive pattern, oscillating between positive and negative amplitudes. The wavelength is the distance over which the wave's shape repeats, while the period is the time taken for one complete oscillation.

(a) If we use equation (15.7) to plot y as a function of x for time t = 0, the curve shows the shape of the string at t = 0.

(b) If we use equation (15.7) to plot y as a function of t for position x = 0, the curve shows the displacement y of the particle at x = 0 as a function of time.

Math summary: This expression calculates the displacement of a wave. It takes time and position as inputs, and scales the cosine of their difference by wave frequency and amplitude to produce the wave's displacement.

Because cosine of negative theta equals cosine theta, we can rewrite the wave function as

Math summary: This expression defines the wave function for a sinusoidal wave, calculating the displacement as a function of location and time. It uses angular frequency and wave speed to determine the wave's behavior in the positive x direction.

The displacement y(x, t) is a function of both the location x of the point and the time t. We could make equation (15.3) more general by allowing for different values of the phase angle, as we did for S.H.M. in Section 14.2, but for now we omit this.

We can rewrite the wave function given by equation (15.3) in several different but useful forms. We can express it in terms of the period T equals 1 over f and the wavelength lambda equals v over f equals 2 times pi times v over omega:

Math summary: This equation calculates the wave function for a sinusoidal wave propagating in the positive x direction. It takes the spatial position and time as inputs, scales them by the wavelength and period respectively, and then calculates the cosine of the result, scaling it by the amplitude to produce the wave function value.

It's convenient to define a quantity k, called the wave number:

Math summary: This expression calculates the wave number. It divides two times pi by the wavelength to produce the wave number.

Substituting lambda equals 2 times pi divided by k and f equals omega divided by 2 times pi into Equation (15.1), v equals lambda times f, gives

Math summary: This expression calculates the angular frequency of a periodic wave. It multiplies the wave's speed by its wave number to determine the angular frequency.

We can then rewrite equation (15.4) as

Math summary: This expression defines the wave function for a sinusoidal wave, showing its amplitude, position, and time components. It describes a wave propagating in the positive x-direction, and also defines the wave number as two pi divided by the wavelength.

Which of these various forms for the wave function y of x and t we use in any specific problem is a matter of convenience. Note that omega has units rad per s, so for unit consistency in Eqs. 15.6 and 15.7 the wave number k must have the units rad per m. (Warning: Some textbooks define the wave number as 1 over lambda rather than 2 times pi over lambda.)

caution Amplitude is independent of wavelength or frequency We learned in Section 15.2 that the frequency and wavelength of a wave are closely related: Increasing the frequency decreases the wavelength, and decreasing the frequency increases the wavelength. The amplitude of a wave, however, does not depend on the frequency or the wavelength. Changing the frequency or wavelength has no effect on the amplitude, and changing the wave amplitude by itself has no effect on the frequency or wavelength.

Graphing the Wave Function

Figure 15.9a graphs the wave function y(x, t) as a function of x for a specific time t. This graph gives the displacement y of a particle from its equilibrium position as a function of the coordinate x of the particle. If the wave is a transverse wave on a string, the graph in figure 15.9a represents the shape of the string at that instant, like a flash photograph of the string. In particular, at time t = 0,

Math summary: This expression calculates the displacement of a wave at time zero. It computes this displacement as a function of position using the cosine of the position scaled by a wave number, with the result scaled by the amplitude.

Figure 15.9b is a graph of the wave function versus time t for a specific coordinate x. This graph gives the displacement y of the particle at x as a function of time; that is, it describes the motion of that particle. At position x = 0,

Math summary: This expression calculates the displacement of a particle at a specific location as a function of time. It takes the amplitude, angular frequency, and time as inputs to compute the displacement using a cosine function.

This is consistent with our original statement about the motion at x = 0, equation (15.2).

Problem-Solving Strategy 15.1 Mechanical Waves

identify the relevant concepts: As always, identify the target variables; these may include mathematical expressions (for example, the wave function for a given situation). Note that wave problems fall into two categories. Kinematics problems, concerned with describing wave motion, involve wave speed v, wavelength lambda (or wave number k), frequency f (or angular frequency omega ), and amplitude A. They may also involve the position, velocity, and acceleration of individual particles in the medium. Dynamics problems also use concepts from Newton's laws. Later in this chapter we'll encounter problems that involve the relationship of wave speed to the mechanical properties of the medium.

Set Up the Problem Using the Following Steps:

1. List the given quantities. Sketch graphs of y versus x (like figure 15.9a) and of y versus t (like figure 15.9b), and label them with known values.

2. Identify useful equations. These may include equation (15.1) (v equals lambda times f), equation (15.6) (omega equals v times k), and Eqs. (15.3), (15.4), and (15.7), which

caution Wave graphs Although they may look the same, Figs. 15.9a and 15.9b are not identical. Figure 15.9a is a picture of the shape of the string at t equals 0, while figure 15.9b is a graph of the displacement y of a particle at x equals 0 as a function of time.

express the wave function in various forms. From the wave function, you can find the value of y at any point (value of 10) and at any time t.

3. If you need to determine the wave speed v and don't know both lambda and f, you may be able to use a relationship between v and the mechanical properties of the system. (In the next section we'll develop this relationship for waves on a string.)

execute the solution: Solve for the unknown quantities using the equations you've identified. To determine the wave function from equation (15.3), (15.4), or (15.7), you must know A and any two of v, λ, and f (or v, k, and ω).

evaluate your answer: Confirm that the values of v, f, and lambda (or v, omega , and k) agree with the relationships given in equation (15.1) or (15.6). If you've calculated the wave function, check one or more special cases for which you can predict the results.

Example 15.2 Wave on a Clothesline

Cousin Throckmorton holds one end of the clothesline taut and wiggles it up and down sinusoidally with frequency 2.00 Hz and amplitude 0.075 m. The wave speed on the clothesline is v equals 12.0 meters per second. At t equals 0 Throcky's end has maximum positive displacement and is instantaneously at rest. Assume that no wave bounces back from the far end. (a) Find the wave amplitude A, angular frequency omega, period T, wavelength lambda, and wave number k. (b) Write a wave function describing the wave. (c) Write equations for the displacement, as a function of time, of Throcky's end of the clothesline and of a point 3.00 m from that end.

identify and set up This is a kinematics problem about the clothesline's wave motion. Throcky produces a sinusoidal wave that propagates along the clothesline, so we can use all of the expressions of this section. In part (a) our target variables are A, omega, T, lambda, and k.

We use the relationships omega equals 2 times pi times f, f equals 1 divided by T, v equals lambda times f, and k equals 2 times pi divided by lambda. In parts (b) and (c) our target "variables" are expressions for displacement, which we'll obtain from an appropriate equation for the wave function. We take the positive x direction to be the direction in which the wave propagates, so either equation (15.4) or (15.7) will yield the desired expression. A photograph of the clothesline at time t equals 0 would look like figure 15.9a, with the maximum displacement at x equals 0 (the end that Throcky holds).

execute (a) The wave amplitude and frequency are the same as for the oscillations of Throcky's end of the clothesline, A equals 0.075 meters and f equals 2.00 Hertz. Hence

Math summary: This calculates angular frequency from frequency. It multiplies two pi radians per cycle by two cycles per second, resulting in approximately twelve point six radians per second.

The period is T equals 1 divided by f equals 0.500 s, and from equation (15.1),

Math summary: This expression calculates wavelength by dividing wave velocity by frequency. The wave velocity of twelve meters per second is divided by the frequency of two inverse seconds, resulting in a wavelength of six meters.

We find the wave number from equation (15.5) or (15.6):

Math summary: This expression calculates the wave number. It divides two pi radians by the wavelength of six meters, resulting in a wave number of one point zero five radians per meter.

or

Math summary: This expression calculates the wave number by dividing angular frequency by velocity. It divides the angular frequency of four pi radians per second by the velocity of twelve meters per second, resulting in a wave number of one point zero five radians per meter.

(b) We write the wave function using equation (15.4) and the values of A, T, and lambda from part (a):

Math summary: This expression calculates the displacement of a wave as a function of position and time. It takes position and time as inputs, scales them by spatial and temporal frequencies, respectively, subtracts the results, and then calculates the cosine of this difference, scaling the final result by an amplitude factor to produce the wave displacement.

We can also get this same expression from equation (15.7) by using the values of omega and k from part (a).

(c) We can find the displacement as a function of time at x = 0 and x = +3.00 m by substituting these values into the wave function from part (b):

Math summary: The expression calculates the displacement as a function of time for two locations on a string. It computes the displacement at each location by evaluating a cosine function with a scaling factor and a time-dependent phase.

evaluate In part (b), the quantity (1.05 rad/m)x - (12.6 radians per second)t is the phase of a point 10 on the string at time t. The two points in part (c) oscillate in S.H.M. with the same frequency and amplitude, but their oscillations differ in phase by (1.05 rad/m)(3.00 m) = 3.15 rad = π radians within rounding error—that is, one half-cycle—because the points are separated by one half-wavelength: λ/2 = (6.00 m)/2 = 3.00 m. Thus, while a graph of y versus t for the point at x = 0 is a cosine curve (like figure 15.9b), a graph of y versus t for the point x = 3.00 m is a negative cosine curve (the same as a cosine curve shifted by one half-cycle).

Using the expression for y(x = 0, t) in part (c), can you show that the end of the string at x = 0 is instantaneously at rest at t = 0, as stated at the beginning of this example? (Hint: Calculate the y-velocity at this point by taking the derivative of y with respect to t.)

keyconcept The wave function of a wave describes the displacement from equilibrium of the wave medium. It gives this displacement at any position in the medium and at any time.

Particle Velocity and Acceleration in a Sinusoidal Wave

From the wave function we can get an expression for the transverse velocity of any particle in a transverse wave. We call this v sub y to distinguish it from the wave propagation speed v. To find the transverse velocity v sub y at a particular point 10, we take the derivative of the wave function y of x and t with respect to t, keeping x constant. If the wave function is

Math summary: This expression calculates the displacement of a point on a wave. It takes position and time as inputs, and outputs the displacement based on amplitude, wave number, and angular frequency.

then

Math summary: This expression calculates the partial derivative of a function y with respect to time t. It outputs the vertical velocity as a function of position x and time t, which is equal to the angular frequency times the amplitude, multiplied by the sine of the wave number times position minus the angular frequency times time.

Definition

Partial Derivative: A derivative of a function with multiple variables, taken with respect to one variable while holding the others constant.

The partial symbol in this expression is a modified d, used to remind us that y of x and t is a function of two variables and that we are allowing only one t to vary. The other x is constant because we are looking at a particular point on the string. This derivative is called a partial derivative. If you haven't yet encountered partial derivatives in your study of calculus, don't fret; it's a simple idea.

Equation (15.9) shows that the transverse velocity of a particle varies with time, as we expect for simple harmonic motion. The maximum particle speed is omega A ; this can be greater than, less than, or equal to the wave speed v, depending on the amplitude and frequency of the wave.

The acceleration of any particle is the second partial derivative of y(x, t) with respect to t:

Math summary: This expression calculates the acceleration of a particle as the second derivative of its displacement with respect to time. The acceleration is found to be equal to a negative constant times the displacement, consistent with simple harmonic motion.

The acceleration of a particle equals -omega sup 2 times its displacement, which is the result we obtained in Section 14.2 for simple harmonic motion.

We can also compute partial derivatives of y of x and t with respect to x, holding t constant. The first partial derivative partial y of x and t, divided by partial x is the slope of the string at point 10 and at time t. The second partial derivative with respect to x tells us the curvature of the string:

Math summary: This expression calculates the second spatial derivative of a wave function. It shows that this second derivative is proportional to the negative square of the wave number, multiplied by the original wave function.

From Eqs. (15.10) and (15.11) and the relationship omega equals v times k we see that

Math summary: This expression equates the ratio of the second derivative of a function with respect to time and the second derivative of the same function with respect to position to the square of the angular frequency divided by the square of the wave number, which is also equal to the square of the wave's velocity.

We've derived equation (15.12) for a wave traveling in the positive x-direction. You can show that the wave function for a sinusoidal wave propagating in the negative x-direction, y of x and t equals A times cosine of k times x plus omega times t, also satisfies this equation.

Definition

Wave Equation: A second-order partial differential equation that describes the propagation of waves in various physical systems.

Equation (15.12), called the wave equation, is one of the most important equations in all of physics. Whenever it occurs, we know that a disturbance can propagate as a wave along the x-axis with wave speed v. The disturbance need not be a sinusoidal wave; we'll see in the next section that any wave on a string obeys equation (15.12), whether the wave is periodic or not. In Chapter 32 we'll find that electric and magnetic fields satisfy the wave equation; the wave speed will turn out to be the speed of light, which will lead us to the conclusion that light is an electromagnetic wave.

Figure 15.10a (next page) shows the transverse velocity v sub y and transverse acceleration a sub y, given by equations (15.9) and (15.10), for several points on a string as a sinusoidal wave passes along it. At points where the string has an upward curvature (partial squared y divided by partial x squared is greater than 0), the acceleration is positive (a sub y equals partial squared y divided by partial t squared is greater than 0); this follows from the wave equation, equation (15.12). For the same reason the acceleration is negative (a sub y equals partial squared y divided by partial t squared is less than 0) at points where the string has a downward curvature (partial squared y divided by partial x squared is less than 0), and the acceleration is zero (a sub y equals partial squared y divided by partial t squared equals 0) at points of inflection where the curvature is zero (partial squared y divided by partial x squared equals 0). Remember that v sub y and a sub y are the transverse velocity and acceleration of points on the string; these points move along the y-direction, not along the propagation direction of the wave. Figure 15.10b shows these motions for several points on the string.

For longitudinal waves, the wave function y(x, t) still measures the displacement of a particle of the medium from its equilibrium position. The difference is that for a longitudinal wave, this displacement is parallel to the x-axis instead of perpendicular to it. We'll discuss longitudinal waves in detail in Chapter 16.

Figure 15.10 (a) Another view of the wave at t equals 0 in Figure 15.9a. The vectors show the transverse velocity v sub y and transverse acceleration a sub y at several points on the string. (b) From t equals 0 to t equals 0.05 times T, a particle at point 1 is displaced to point 1 prime, a particle at point 2 is displaced to point 2 prime, and so on.

Figure 15.10 summary: The figure is a wave diagram that shows the motion of particles in a transverse wave. The diagram illustrates the vertical velocity and acceleration of different points along the wave at a specific time. The figure also shows the wave's displacement at two different times. The velocity of a particle is greater when closer to the equilibrium position, while acceleration is greater at the maximum displacement. The wave propagates along the x-axis as time increases.

test your understanding of section 15.3 Figure 15.8 shows a sinusoidal wave of period T on a string at times 0, 1 eighth times T, 2 eighths times T, 3 eighths times T, 4 eighths times T, 5 eighths times T, 6 eighths times T, 7 eighths times T, and T. (a) At which time is point A on the string moving upward with maximum speed? (b) At which time does point B on the string have the greatest upward acceleration? (c) At which time does point C on the string have a downward acceleration and a downward velocity?

Answer

C has an upward displacement and is moving downward at t equals five eighths times T. wave depicted in figure 15.8. (a) A particle in S.H.M. has its maximum speed when it is passing through the equilibrium position (y = 0 in figure 15.8). The particle at point A is moving upward through this position at t = 2 over 8 T . (b) In vertical S.H.M. the greatest upward acceleration occurs when a particle is at its maximum downward displacement. This occurs for the particle at point B at t = 4 over 8 T . (c) A particle in vertical S.H.M. has a downward acceleration when its displacement is upward. The particle at (a) 2 over 8 times T, (b) 4 over 8 times T, (c) 5 over 8 times T. Since the wave is sinusoidal, each point on the string oscillates in simple harmonic motion (S.H.M.). Hence we can apply all of the ideas from Chapter 14 about S.H.M. to the

15.4 Speed of a Transverse Wave

One of the key properties of any wave is the wave speed. Light waves in air have a much greater speed of propagation than do sound waves in air ( 3.00 times 10 sup 8 meters per second versus 344 meters per second ); that's why you see the flash from a bolt of lightning before you hear the clap of thunder. In this section we'll see what determines the speed of propagation of one particular kind of wave: transverse waves on a string. The speed of these waves is important to understand because it is an essential part of analyzing stringed musical instruments, as we'll discuss later in this chapter. Furthermore, the speeds of many kinds of mechanical waves turn out to have the same basic mathematical expression as does the speed of waves on a string.

What determines the speed of transverse waves on a string are the tension in the string and its mass per unit length (also called linear mass density). Increasing the tension also increases the restoring forces that tend to straighten the string when it is disturbed, thus increasing the wave speed. Increasing the mass per unit length makes the motion more sluggish, and so decreases the wave speed.

We'll develop the exact relationship among wave speed, tension, and mass per unit length by two different methods. The first is simple in concept and considers a specific wave shape; the second is more general but also more formal.

Wave Speed on a String: First Method

We consider a perfectly flexible string (Fig. 15.11). In the equilibrium position the tension is F and the linear mass density (mass per unit length) is mu . (When portions of the string are displaced from equilibrium, the mass per unit length decreases a little, and the tension increases a little.) We ignore the weight of the string so that when the string is at rest in the equilibrium position, the string forms a perfectly straight line as in figure 15.11a.

Figure 15.11 summary: The figure illustrates the propagation of a transverse wave on a string. It depicts a string under tension, initially at equilibrium, and then shows a part of the string in motion as a wave propagates through it. The wave's propagation involves the string being displaced vertically, with different segments moving upward with varying velocities, while the disturbance itself propagates at a specific wave speed. The string's behavior demonstrates how a transverse wave is formed and travels along the string due to applied forces and resulting motion.

Starting at time t = 0, we apply a constant upward force F y at the left end of the string. We might expect that the end would move with constant acceleration; that would happen if the force were applied to a point mass. But here the effect of the force F y is to set successively more and more mass in motion. The wave travels with constant speed v, so the division point P between moving and nonmoving portions moves with the same constant speed v (Fig. 15.11b).

Figure 15.11b shows that all particles in the moving portion of the string move upward with constant velocity v y , not constant acceleration. To see why this is so, we note that the impulse of the force F y up to time t is F y t . According to the impulse–momentum theorem (see Section 8.1), the impulse is equal to the change in the total transverse component of momentum of the moving part of the string. Because the system started with zero transverse momentum, this is equal to the total transverse momentum mv y at time t :

Transverse impulse = Transverse momentum

Math summary: This equation equates transverse impulse and transverse momentum. It states that the product of the transverse force and time equals the product of mass and transverse velocity.

The total momentum thus must increase proportionately with time. But since the division point P moves with constant speed, the length of string that is in motion and hence the total mass m in motion are also proportional to the time t that the force has been acting. So the change of momentum must be associated entirely with the increasing amount of mass in motion, not with an increasing velocity of an individual mass element. That is, mv sub y changes because m, not v sub y , changes.

At time t, the left end of the string has moved up a distance v sub y times t, and the boundary point P has advanced a distance v times t. The total force at the left end of the string has components F and F sub y. Why F? There is no motion in the direction along the length of the string, so there is no unbalanced horizontal force. Therefore F, the magnitude of the horizontal component, does not change when the string is displaced. In the displaced position the tension is, open parenthesis, F squared plus F sub y squared, close parenthesis, to the power of 2, close parenthesis, times one half; this is greater than F, so the string stretches somewhat.

To derive an expression for the wave speed v, we note that in figure 15.11b the right triangle whose vertex is at P, with sides v sub y and vt, is similar to the right triangle whose vertex is at the position of the hand, with sides F sub y and F. Hence

Math summary: First, the ratio of the vertical force component to the total force is set equal to the ratio of the vertical velocity component times time to the total velocity times time. Then, the vertical force component is calculated as the total force multiplied by the ratio of the vertical velocity component to the total velocity.

and

Math summary: This equation calculates transverse impulse. It multiplies the force component in the y direction by time, which is equivalent to multiplying the total force by the ratio of the y component of velocity to the total velocity, and then by time.

Figure 15.12 These transmission cables have a relatively large amount of mass per unit length (mu) and a low tension (F). If the cables are disturbed—say, by a bird landing on them—transverse waves will travel along them at a slow speed v equals square root of F divided by mu.

Figure 15.12 summary: The table presents the components of the wave equation, highlighting the relationship between the second partial derivative of the wave function with respect to both position and time, and also the wave speed.

The mass m of the moving portion of the string is the product of the mass per unit length mu and the length v times t, or mu times v times t. The transverse momentum is the product of this mass and the transverse velocity v sub y:

Math summary: This expression calculates transverse momentum. It multiplies the mass of a moving string portion by its transverse velocity to determine the momentum.

Substituting these into equation (15.13), we obtain We solve this for the wave speed v:

Math summary: This expression calculates the speed of a transverse wave on a string. It takes the tension in the string and the mass per unit length as inputs, divides the tension by the mass per unit length, and then calculates the square root of the result to determine the wave speed.

Equation (15.14) confirms that the wave speed v increases when the tension F increases but decreases when the mass per unit length mu increases (Fig. 15.12).

Note that v sub y does not appear in equation (15.14); thus the wave speed doesn't depend on v sub y . Our calculation considered only a very special kind of pulse, but we can consider any shape of wave disturbance as a series of pulses with different values of v sub y . So even though we derived equation (15.14) for a special case, it is valid for any transverse wave motion on a string, including the sinusoidal and other periodic waves we discussed in Section 15.3. Note also that the wave speed doesn't depend on the amplitude or frequency of the wave, in accordance with our assumptions in Section 15.3.

Wave Speed on a String: Second Method

Here is an alternative derivation of Equation fifteen point fourteen. If you aren't comfortable with partial derivatives, it can be omitted. We apply Newton's second law, the sum of the vector F equals m times vector a, to a small segment of string whose length in the equilibrium position is delta x, Figure fifteen point thirteen. The mass of the segment is m equals mu times delta x. The x components of the forces have equal magnitude F and add to zero because the motion is transverse and there is no component of acceleration in the x direction. To obtain F sub one y and F sub two y, we note that the ratio F sub one y divided by F is equal in magnitude to the slope of the string at point 10 and that F sub two y divided by F is equal to the slope at point 10 plus delta x. Taking proper account of signs, we find

Math summary: The expression calculates ratios of force components to total force using partial derivatives of a function. Specifically, it relates the ratio of the y-component of force at a point to the total force with the negative partial derivative of y with respect to x at that point, and similarly relates another force ratio to the partial derivative at a slightly different point.

Figure 15.13 summary: The figure is a free-body diagram that illustrates the forces acting on a small segment of a string. The diagram shows the forces at each end of the string segment, resolved into horizontal and vertical components. The force at each end is tangent to the string at the point of application. The diagram indicates that while there can be a net vertical force, the net horizontal force on the segment is zero, implying transverse motion.

The notation reminds us that the derivatives are evaluated at points 10 and x + Delta x , respectively. From equation (15.15) we find that the net y-component of force is

Math summary: This expression calculates the net y component of force as the sum of two force components. It equals the product of a force and the difference between two partial derivatives of y with respect to x, evaluated at x plus delta x and x.

We now equate F sub y from Equation fifteen point sixteen to the mass mu times delta x times the y component of acceleration partial squared y over partial t squared:

Math summary: This equation calculates a relationship between the change in the partial derivative of y with respect to x and the second partial derivative of y with respect to time. The difference in the partial derivative of y with respect to x is scaled and equated to mass times change in x times the second partial derivative of y with respect to time.

or, dividing Equation fifteen point seventeen by F times delta x

Math summary: This expression calculates the relationship between the change in the rate of change of y with respect to x and the rate of change of y with respect to time. It equates the change in the partial derivative of y with respect to x, divided by a small change in x, to a scaled second partial derivative of y with respect to time.

We now take the limit as delta x approaches 0. In this limit, the left side of equation (15.18) becomes the derivative of the partial derivative of y with respect to x with respect to x at constant t that is, the second partial derivative of y with respect to x.

Math summary: This equation relates the second partial derivative of a function y with respect to x to the second partial derivative of y with respect to time t. It states that the second derivative with respect to x equals the second derivative with respect to time multiplied by a scaling factor of mu divided by F.

Now, equation (15.19) has exactly the same form as the wave equation, equation (15.12), that we derived at the end of Section 15.3. That equation and equation (15.19) describe the very same wave motion, so they must be identical. Comparing the two equations, we see that for this to be so, we must have

Math summary: This expression calculates the velocity. It takes the square root of the tension divided by the linear density to produce the velocity.

which is the same expression as equation (15.14).

In going through this derivation, we didn't make any special assumptions about the shape of the wave. Since our derivation led us to rediscover equation (15.12), the wave equation, we conclude that the wave equation is valid for waves on a string that have any shape.

The Speed of Mechanical Waves

Equation (15.14) gives the wave speed for only the special case of mechanical waves on a stretched string or rope. Remarkably, it turns out that for many types of mechanical waves, including waves on a string, the expression for wave speed has the same general form:

Math summary: This formula calculates wave speed by taking the square root of a fraction. The fraction is the restoring force divided by the inertia resisting the return to equilibrium; this result is the wave speed.

To interpret this expression, let's look at the now-familiar case of waves on a string. The tension F in the string plays the role of the restoring force; it tends to bring the string back to its undisturbed, equilibrium configuration. The mass of the string—or, more properly, the linear mass density mu—provides the inertia that prevents the string from returning instantaneously to equilibrium. Hence we have v = square root of F/mu for the speed of waves on a string.

In Chapter 16 we'll see a similar expression for the speed of sound waves in a gas. Roughly speaking, the gas pressure provides the force that tends to return the gas to its undisturbed state when a sound wave passes through. The inertia is provided by the density, or mass per unit volume, of the gas.

bio Application Eating and

Transverse Waves Swallowing food causes peristalsis, in which a transverse wave propagates down your esophagus. The wave is a radial contraction of the esophagus that pushes the bolus (the mass of swallowed food) toward the stomach. Unlike the speed of waves on a uniform string, the speed of this peristaltic wave is not constant: It averages about 3 centimeters/s in the upper esophagus, about 5 centimeters/s in the mid-esophagus, and about 2.5 centimeters/s in the lower esophagus.

Image summary: The image is a diagram that illustrates the motion of a peristaltic wave. The diagram depicts a bolus moving through a tube-like structure, presumably the esophagus, from the mouth towards the stomach. The bolus is shown progressing through the structure, with the structure constricting behind it and relaxing in front of it. This peristaltic movement propels the bolus forward. The diagram suggests that the peristaltic wave is responsible for moving the bolus along the digestive tract.

Example 15.3 Calculating Wave Speed

One end of a 2.00 kilograms rope is tied to a support at the top of a mine shaft 80.0 m deep (Fig. 15.14). The rope is stretched taut by a 20.0 kilograms box of rocks attached at the bottom. (a) A geologist at the bottom of the shaft signals to a colleague at the top by jerking the rope sideways. What is the speed of a transverse wave on the rope? (b) If a point on the rope is in transverse S.H.M. with f = 2.00 Hz , how many cycles of the wave are there in the rope's length?

Figure 15.14 summary: The figure is an illustration depicting a person at the top of a cliff lowering a rope to another person at the bottom of the cliff. The illustration provides information about the mass of the rope and the mass of the samples being lifted. The image suggests a scenario involving lifting samples from the bottom of a cliff using a rope, with the mass of the rope and the samples being important factors.

identify and set up In part (a) we can find the wave speed (our target variable) by using the dynamic relationship v = square root of F/mu . In part (b) we find the wavelength from the kinematic relationship v = flambda ; from that we can find the target variable, the number of wavelengths that fit into the rope's 80.0 m length. We'll assume that the rope is massless (even though its weight is 10% that of the box), so that the box alone provides the tension in the rope.

With Variation Problems

execute (a) The tension in the rope due to the box is

Math summary: This expression calculates the force due to gravity on a box. It multiplies the mass of the box by the acceleration due to gravity to produce the force in Newtons.

and the rope's linear mass density is

Math summary: This expression calculates the linear mass density of a rope. It divides the rope's total mass of two kilograms by its length of eighty meters, resulting in a linear mass density of 0.025 kilograms per meter.

Figure 15.14 Using transverse waves to send signals along a vertical rope.

Hence, from equation (15.14), the wave speed is

Math summary: This expression calculates the wave speed by taking the square root of the tension force divided by the linear density. The tension force of 196 newtons is divided by the linear density of 0.0250 kilograms per meter, and then the square root of the result is taken, yielding a wave speed of 88.5 meters per second.

(b) From equation (15.1), the wavelength is

Math summary: This calculates the wavelength of a wave. It divides the wave's velocity, which is 88.5 meters per second, by its frequency, which is 2.00 inverse seconds, resulting in a wavelength of 44.3 meters.

There are 80.0 meters divided by 44.3 meters equals 1.81 wavelengths (that is, cycles of the wave) in the rope. evaluate Because of the rope's weight, its tension is greater at the top than at the bottom. Hence both the wave speed and the wavelength increase as a wave travels up the rope. If you take account of this, can you verify that the wave speed at the top of the rope is 92.9 meters per second?

keyconcept The speed of a transverse wave on a string is determined by the tension in the string and the string's linear mass density (mass per unit length). For a given string, the wave speed is independent of frequency or wavelength.

test your understanding of section 15.4 The six strings of a guitar are the same length and under nearly the same tension, but they have different thicknesses. On which string do waves travel the fastest? (1) The thickest string; (2) the thinnest string; (3) the wave speed is the same on all strings.

Answer

(2) The relationship v = square root of F/mu says that the wave speed is greatest on the string with the smallest linear mass density. This is the thinnest string, which has the smallest amount of mass m and hence the smallest linear mass density mu = m/L (all strings are the bio application Surface Waves and the Swimming Speed of Ducks When a duck swims, it necessarily produces waves on the surface of the water. The faster the duck swims, the larger the wave amplitude and the more power the duck must supply to produce these waves. The maximum power available from their leg muscles limits the maximum swimming speed of ducks to only about 0.7 meters per second (2.5 kilometers/h = 1.6 mi/h).

Figure 15.15 summary: The figure contains two diagrams. The first diagram illustrates a point on a string that is experiencing a wave moving from left to right. The second diagram shows the force components exerted on a part of the string to the right of a specific point, caused by the part of the string to the left of that point. The diagrams suggest that the wave motion causes forces on the string, which can be broken down into horizontal and vertical components.

15.5 Energy in Wave Motion

Every wave motion has energy associated with it. The energy we receive from sunlight and the destructive effects of ocean surf and earthquakes bear this out. To produce any of the wave motions we have discussed in this chapter, we have to apply a force to a portion of the wave medium; the point where the force is applied moves, so we do work on the system. As the wave propagates, each portion of the medium exerts a force and does work on the adjoining portion. In this way a wave can transport energy from one region of space to another.

As an example, let's look again at transverse waves on a string. How is energy transferred from one portion of the string to another? Picture a wave traveling from left to right (the positive x-direction) past a point a on the string (Fig. 15.15a). The string to the left of point a exerts a force on the string to the right of it, and vice versa. In figure 15.15b we show the components F x and F y of the force that the string to the left of a exerts on the string to the right of a. As in Figs. 15.11 and 15.13, the magnitude of the horizontal component F x equals the tension F in the undisturbed string. Note that F y/F is equal to the negative of the slope of the string at a, and this slope is also given by partial y/partial x . Putting these together, we have

Math summary: This expression calculates the vertical force at a specific point and time on a string. It multiplies the negative of the string tension by the rate of change of the vertical displacement with respect to the horizontal position.

We need the negative sign because F y is negative when the slope is positive (as in figure 15.15b). We write the vertical force as F y(x, t) as a reminder that its value may be different at different points along the string and at different times.

When point a moves in the y direction, the force F sub y does work on this point and therefore transfers energy into the part of the string to the right of a. The corresponding power P, rate of doing work, at the point a is the transverse force F sub y of x and t at a times the transverse velocity v sub y of x and t equals the partial derivative of y of x and t with respect to t of that point:

Math summary: This expression calculates the power transmitted along a string as the product of the transverse force and the transverse velocity. It is also equivalent to negative the product of the tension, the rate of change of displacement with respect to position, and the rate of change of displacement with respect to time.

This power is the instantaneous rate at which energy is transferred along the string at position x and time t. Note that energy is transferred only at points where the string has a nonzero slope ( partial y/partial x is nonzero), so that the tension force has a transverse component, and where the string has a nonzero transverse velocity ( partial y/partial t is nonzero) so that the transverse force can do work.

Equation (15.21) is valid for any wave on a string, sinusoidal or not. For a sinusoidal wave with wave function given by equation (15.7), we have

Math summary: The expression calculates the power of a sinusoidal wave on a string. It takes the force, wave number, angular frequency, and amplitude as inputs, and outputs the instantaneous power as a function of position and time, proportional to the square of the sine function.

By using the relationships omega equals v times k and v squared equals F divided by mu, we can also express equation (15.22) in the alternative form

Math summary: This expression calculates the instantaneous power of a sinusoidal wave. It takes the square root of the product of linear density and tension, multiplies it by the square of the angular frequency and the square of the amplitude, and then multiplies the result by the square of the sine of the phase, which is the wave number times position minus the angular frequency times time.

The sin sup 2 function is never negative, so the instantaneous power in a sinusoidal wave is either positive (so that energy flows in the positive x-direction) or zero (at points where there is no energy transfer). Energy is never transferred in the direction opposite to the direction of wave propagation (Fig. 15.16).

Figure 15.16 summary: The figure is a line plot. It illustrates the relationship between wave power and time at a specific coordinate. The wave power fluctuates periodically over time, reaching a maximum value and decreasing to zero. The average power is exactly half of the maximum power.

The maximum value of the instantaneous power P of x and t occurs when the sine squared function has the value unity:

Math summary: This expression calculates the maximum power. It takes the square root of the product of two inputs, multiplies it by the square of another input, and then by the square of a final input to produce the maximum power.

The average value of the sine squared function, averaged over any whole number of cycles, is 1 over 2. Hence we see from equation (15.23) that the average power P subscript av is just one-half the maximum instantaneous power P subscript max (Fig. 15.16):

Math summary: This calculates the average power of a sinusoidal wave on a string. It multiplies one half by the square root of the product of the string's linear mass density and tension, then multiplies by the square of the wave's angular frequency and the square of the wave's amplitude.

The average rate of energy transfer is proportional to the square of the amplitude and to the square of the frequency. This proportionality is a general result for mechanical waves of all types, including seismic waves (see the photo that opens this chapter). For a mechanical wave, the rate of energy transfer quadruples if the frequency is doubled (for the same amplitude) or if the amplitude is doubled (for the same frequency).

Electromagnetic waves turn out to be a bit different. While the average rate of energy transfer in an electromagnetic wave is proportional to the square of the amplitude, just as for mechanical waves, it is independent of the value of omega . Figure 15.16 The instantaneous power P(x, t) in a sinusoidal wave as given by equation (15.23), shown as a function of time at coordinate x = 0. The power is never negative, which means that energy never flows opposite to the direction of wave propagation.

Example 15.4 Power in a Wave

(a) In Example 15.2 (Section 15.3), at what maximum rate does Throcky put energy into the clothesline? That is, what is his maximum instantaneous power? The linear mass density of the clothesline is mu equals 0.250 kilograms per meter, and Throcky applies tension F equals 36.0 Newtons. (b) What is his average power? (c) As Throcky tires, the amplitude decreases. What is the average power when the amplitude is 7.50 millimeters?

With Variation Problems

identify and set up In part (a) our target variable is the maximum instantaneous power P sub max , while in parts (b) and (c) it is the average power. For part (a) we'll use equation (15.24), and for parts (b) and (c) we'll use equation (15.25); Example 15.2 gives us all the needed quantities. execute (a) From equation (15.24),

Math summary: This expression calculates the maximum instantaneous power. It takes the square root of the product of the linear mass density and the tension, then multiplies it by the square of the angular frequency and the square of the amplitude, resulting in the maximum instantaneous power.

(b) From Eqs. (15.24) and (15.25), the average power is one-half of the maximum instantaneous power, so

Math summary: This calculates the average power. It takes one half of the maximum power, which is 2.66 watts, to produce an average power of 1.33 watts.

(c) The new amplitude is 1 over 10 of the value we used in parts (a) and (b). From equation (15.25), the average power is proportional to A sup 2 , so the new average power is

Math summary: This calculation determines the average power. It squares one tenth, multiplies the result by one point thirty-three watts, and obtains an average power of thirteen point three milliwatts.

evaluate Equation (15.23) shows that P sub max occurs when sine squared of (k times x minus omega times t) equals 1. At any given position x, this happens twice per period of the wave—once when the sine function is equal to positive 1, and once when it's equal to negative 1. The minimum instantaneous power is zero; this occurs when sine squared of (k times x minus omega times t) equals 0, which also happens twice per period. Can you confirm that the given values of mu and F give the wave speed mentioned in Example 15.2?

keyconcept The power (energy per second) carried by a wave on a string depends on the string tension and the linear mass density, as well as the frequency and amplitude of the wave. If the wave is sinusoidal, the average power equals one-half of the maximum instantaneous power.

Wave Intensity

Waves on a string carry energy in one dimension (along the direction of the string). But other types of waves, including sound waves in air and seismic waves within the earth, carry energy across all three dimensions of space. For waves of this kind, we define the intensity (denoted by 1) to be the time average rate at which energy is transported by the wave, per unit area, across a surface perpendicular to the direction of propagation. Intensity I is average power per unit area and is usually measured in watts per square meter (W/m²).

If waves spread out equally in all directions from a source, the intensity at a distance r from the source is inversely proportional to r ^2 (Fig. 15.17). This result, called the inverse-square law for intensity, follows directly from energy conservation. If the power output of the source is P, then the average intensity I 1 through a sphere with radius r 1 and surface area 4pi r 1 ^2 is

Math summary: This equation calculates the intensity at a specific distance from a wave source. It divides the power output of the source by the surface area of a sphere with a radius equal to that distance.

Figure 15.17 summary: The figure is an illustration that depicts the relationship between distance from a wave source and wave intensity. The figure illustrates a wave source emitting waves outwards. It shows two distances from the source, one closer and one farther away. The intensity at the closer distance is higher, while the intensity at the farther distance is lower. The figure suggests that as the distance from the wave source increases, the wave intensity decreases because the same power is distributed over a larger area.

A similar expression gives the average intensity I sub 2 through a sphere with a different radius r sub 2 . If no energy is absorbed between the two spheres, the power P must be the same for both, and

Math summary: This equation equates two power calculations. Each side of the equation calculates power as the product of four pi, the square of a radius, and an intensity value at that radius.

Inverse-square law for intensity: Intensity is inversely proportional to the square of the distance from source.

Intensity at point 1 goes to I sub 1

Example 15.5 The Inverse-Square Law

With Variation Problems

A siren on a tall pole radiates sound waves uniformly in all directions. At a distance of 15.0 m from the siren, the sound intensity is 0.250 W/m². At what distance is the intensity 0.010 W/m²?

identify and set up Because sound is radiated uniformly in all directions, we can use the inverse-square law, Equation fifteen point twenty six. At r sub 1 equals fifteen point zero meters the intensity is I sub 1 equals zero point two five zero watts per meter squared, and the target variable is the distance r sub 2 at which the intensity is I sub 2 equals zero point zero one zero watts per meter squared.

execute We solve Equation fifteen point twenty six for r sub two:

Math summary: This expression calculates a new radius. It takes an initial radius and multiplies it by the square root of the ratio of the initial intensity to the new intensity, resulting in a final radius.

evaluate As a check on our answer, note that r sub 2 is five times greater than r sub 1. By the inverse-square law, the intensity I sub 2 should be 1 divided by 5 squared equals 1 divided by 25 as great as I sub 1, and indeed it is.

By using the inverse-square law, we've assumed that the sound waves travel in straight lines away from the siren. A more realistic solution, which is beyond our scope, would account for the reflection of sound waves from the ground.

keyconcept If waves emitted from a source spread out equally in all directions, the wave intensity is inversely proportional to the square of the distance from the source.

test your understanding of section 15.5 Four identical strings each carry a sinusoidal wave of frequency 10 Hz. The string tension and wave amplitude are different for different strings. Rank the following strings in order from highest to lowest value of the average wave power: (1) tension 10 N, amplitude 1.0 millimeters; (2) tension 40 N, amplitude 1.0 millimeters; (3) tension 10 N, amplitude 4.0 millimeters; (4) tension 20 N, amplitude 2.0 millimeters.

Answer

times greater. omega equals 2 times pi times f. Hence the average wave power for each string is proportional to the square root of the string tension F and the square of the amplitude A. Compared to string (i), the average power in each string is (2) square root of 4 equals 2 times greater; (3) 4 squared equals 16 times greater; and (4) square root of 2 times 2 squared equals 4 times square root of 2 (3), (4), (2), (i) Equation (15.25) says that the average power in a sinusoidal wave on a string is P sub av equals one half times square root of mu times F, times omega squared times A squared. All four strings are identical, so all have the same mass, length, and linear mass density mu. The frequency f is the same for each wave, as is the angular frequency

15.6 Wave Interference, Boundary Conditions, and Superposition

Up to this point we've been discussing waves that propagate continuously in the same direction. But when a wave strikes the boundaries of its medium, all or part of the wave is reflected. When you yell at a building wall or a cliff face some distance away, the sound wave is reflected from the rigid surface and you hear an echo. When you flip the end of a rope whose far end is tied to a rigid support, a pulse travels the length of the rope and is reflected back to you. In both cases, the initial and reflected waves overlap in the same region of the medium. We use the term interference to refer to what happens when two or more waves pass through the same region at the same time.

As a simple example of wave reflections and the role of the boundary of a wave medium, let's look again at transverse waves on a stretched string. What happens when a wave pulse or a sinusoidal wave arrives at the end of the string?

If the end is fastened to a rigid support as in figure 15.18, it is a fixed end that cannot move. The arriving wave exerts a force on the support (drawing 4 in figure 15.18); the reaction to this force, exerted by the support on the string, “kicks back” on the string and sets up a reflected pulse or wave traveling in the reverse direction (drawing 7). The reflected pulse moves in the opposite direction from the initial, or incident, pulse, and its displacement is also opposite.

Figure 15.18 summary: The figure illustrates the reflection of a wave pulse at a fixed end of a string over time. The wave pulse travels along the string toward the fixed end. Upon reaching the fixed end, the string exerts an upward force on the wall, and in response, the wall exerts a downward reaction force on the string. The wave pulse inverts and travels in the opposite direction after reflecting from the fixed end.

The opposite situation from an end that is held stationary is a free end, one that is perfectly free to move in the direction perpendicular to the length of the string. For example, the string might be tied to a light ring that slides on a frictionless rod perpendicular to the string, as in figure 15.19. The ring and rod maintain the tension but exert no transverse force. When a wave arrives at this free end, the ring slides along the rod.

Figure 15.19 summary: The figure illustrates the reflection of a wave pulse at the free end of a string over time. The figure shows the wave pulse approaching a vertical rod, which exerts no transverse forces on the string. The wave pulse is then shown reflecting from the free end. The wave pulse reflects without inverting. Therefore, the wave pulse maintains its orientation after reflection.

The ring reaches a maximum displacement, and both it and the string come momentarily to rest, as in drawing 4 in figure 15.19. But the string is now stretched, giving increased tension, so the free end of the string is pulled back down, and again a reflected pulse is produced (drawing 7). As for a fixed end, the reflected pulse moves in the opposite direction from the initial pulse, but now the direction of the displacement is the same as for the initial pulse. The conditions at the end of the string, such as a rigid support or the complete absence of transverse force, are called boundary conditions.

The formation of the reflected pulse is similar to the overlap of two pulses traveling in opposite directions. Figure 15.20 (next page) shows two pulses with the same shape, one inverted with respect to the other, traveling in opposite directions. As the pulses overlap and pass each other, the total displacement of the string is the algebraic sum of the displacements at that point in the individual pulses.

Figure 15.20 summary: This figure depicts a sequence of snapshots illustrating the overlap of two wave pulses traveling in opposite directions. One pulse is right-side up, while the other is inverted. As the pulses approach, they begin to overlap, resulting in a superposition. At the point of complete overlap, the pulses cancel each other out. After passing through each other, the pulses continue to propagate in their original directions with their original shapes, showing destructive interference.

Because these two pulses have the same shape, the total displacement at point O in the middle of the figure is zero at all times. Thus the motion of the left half of the string would be the same if we cut the string at point O, threw away the right side, and held the end at O fixed. The two pulses on the left side then correspond to the incident and reflected pulses, combining so that the total displacement at O is always zero. For this to occur, the reflected pulse must be inverted relative to the incident pulse, just as for reflection from the fixed end in figure 15.18.

As the pulses overlap, the displacement of the string at any point is the algebraic sum of the displacements due to the individual pulses.

Figure 15.21 shows two pulses with the same shape, traveling in opposite directions but not inverted relative to each other. The displacement at point O in the middle of the figure is not zero, but the slope of the string at this point is always zero. According to equation (15.20), this corresponds to the absence of any transverse force at this point.

Figure 15.21 summary: The figure is a depiction of the overlap of two wave pulses traveling in opposite directions. The image displays several snapshots in time, showing the progression of the pulses as they approach, interact, and then separate. As the pulses move closer, their amplitudes combine to create a larger amplitude where they overlap. After the interference, the pulses continue to propagate in their original directions, retaining their initial shapes and amplitudes, as if the interaction did not occur.

In this case the motion of the left half of the string would be the same as if we cut the string at point O and attached the end to a frictionless sliding ring (Fig. 15.19) that maintains tension without exerting any transverse force. In other words, this situation corresponds to reflection of a pulse at a free end of a string at point O. In this case the reflected pulse is not inverted.

The Principle of Superposition

Definition

Principle of Superposition: The principle stating that when two or more waves overlap, the resulting displacement is the sum of the individual displacements.

Combining the displacements of the separate pulses at each point to obtain the actual displacement is an example of the principle of superposition: When two waves overlap, the actual displacement of any point on the string at any time is obtained by adding the displacement the point would have if only the first wave were present and the displacement it would have if only the second wave were present. In other words, the wave function y(x, t) for the resulting motion is obtained by adding the two wave functions for the two separate waves:

Wave functions of two overlapping waves Principle of superposition:

Math summary: This expression calculates the superposition of two wave functions. It sums the first wave function and the second wave function to produce a combined wave function.

Mathematically, this additive property of wave functions follows from the form of the wave equation, equation (15.12) or (15.19), which every physically possible wave function must satisfy. Specifically, the wave equation is linear; that is, it contains the function y (x,t) only to the first power (there are no terms involving y(x,t)^{2} , y(x,t)^{1/2} , etcetera). As a result, if any two functions y sub 1 (x,t) and y sub 2 (x,t) satisfy the wave equation separately, their sum y sub 1 (x,t) + y sub 2 (x,t) also satisfies it and is therefore a physically possible motion. Because this principle depends on the linearity of the wave equation and the corresponding linear-combination property of its solutions, it is also called the principle of linear superposition. For some physical systems, such as a medium that does not obey Hooke's law, the wave equation is not linear; this principle does not hold for such systems.

The principle of superposition is of central importance in all types of waves. When a friend talks to you while you are listening to music, you can distinguish the speech and the music from each other. This is precisely because the total sound wave reaching your ears is the algebraic sum of the wave produced by your friend's voice and the wave produced by the speakers of your stereo. If two sound waves did not combine in this simple linear way, the sound you would hear in this situation would be a hopeless jumble. Superposition also applies to electromagnetic waves (such as light). test your understanding of section 15.6 Figure 15.22 shows two wave pulses with different shapes traveling in different directions along a string. Make a series of sketches like figure 15.21 showing the shape of the string as the two pulses approach, overlap, and then pass each other. answer Figure 15.22 Two wave pulses with different shapes.

Figure 15.22 summary: The figure illustrates the overlap of two wave pulses traveling in opposite directions. The figure shows the progression of the wave pulses over time. As the wave pulses approach each other, they begin to overlap. The amplitude of the resulting wave is greater during the period of overlap than the individual waves. After the wave pulses pass through each other, they continue to propagate in their original directions with their original shapes.

15.7 Standing Waves on a String

We've looked at the reflection of a wave pulse on a string when it arrives at a boundary point (either a fixed end or a free end). Now let's consider what happens when a sinusoidal wave on a string is reflected by a fixed end. We'll again approach the problem by considering the superposition of two waves propagating through the string, one representing the incident wave and the other representing the wave reflected at the fixed end.

Figure 15.23 shows a string that is fixed at its left end. Its right end is moved up and down in simple harmonic motion to produce a wave that travels to the left; the wave reflected from the fixed end travels to the right. The resulting motion when the two waves combine no longer looks like two waves traveling in opposite directions.

Figure 15.23 summary: The figure is a sequence of images that demonstrate the overlap of two wave pulses, both right side up, traveling in opposite directions. The images show the wave pulses approaching each other, overlapping to create a larger amplitude, and then separating as they continue to move in opposite directions. The superposition principle is evident as the amplitudes of the waves combine during the overlap. The wave pulses maintain their original shape and direction of propagation after they have completely passed through each other.

3 Definitions

Definition 1: Standing Wave: A wave pattern formed by the interference of two waves traveling in opposite directions, characterized by fixed points (nodes) and points of maximum amplitude (antinodes).

Definition 2: Nodes: Points in a standing wave where the amplitude is always zero.

Definition 3: Antinodes: Points in a standing wave where the amplitude is maximum.

The string appears to be subdivided into segments, as in the time-exposure photographs of Figs. 15.23a, 15.23b, 15.23c, and 15.23d. Figure 15.23e shows two instantaneous shapes of the string in figure 15.23b. Let's compare this behavior with the waves we studied in Sections 15.1 through 15.5. In a wave that travels along the string, the amplitude is constant and the wave pattern moves with a speed equal to the wave speed. Here, instead, the wave pattern remains in the same position along the string and its amplitude fluctuates. There are particular points called nodes (labeled N in figure 15.23e) that never move at all. Midway between the nodes are points called antinodes (labeled A in figure 15.23e) where the amplitude of motion is greatest. Because the wave pattern doesn't appear to be moving in either direction along the string, it is called a standing wave. (To emphasize the difference, a wave that does move along the string is called a traveling wave.)

The principle of superposition explains how the incident and reflected waves combine to form a standing wave. In figure 15.24 (next page) the red curves show a wave traveling to the left. The blue curves show a wave traveling to the right with the same propagation speed, wavelength, and amplitude.

Figure 15.24 summary: The figure is a combination of plots depicting the behavior of a string that is two wavelengths long. The figure illustrates the standing wave pattern on the string, showing nodes and antinodes. The string's shape is shown at different instants, demonstrating how the waves interfere constructively and destructively, resulting in maximum displacement and cancellation.

The waves are shown at nine instants, 1 over 16 of a period apart. At each point along the string, we add the displacements (the values of y) for the two separate waves; the result is the total wave on the string, shown in gold.

At certain instants, such as t = 1 over 4T , the two wave patterns are exactly in phase with each other, and the shape of the string is a sine curve with twice the amplitude of either individual wave. At other instants, such as t = 1 over 2T , the two waves are exactly out of phase with each other, and the total wave at that instant is zero. The resultant displacement is always zero at those places marked N at the bottom of figure 15.24. These are the nodes. At a node the displacements of the two waves in red and blue are always equal and opposite and cancel each other out.

Image summary: The image shows a depiction of standing waves on a string. The string demonstrates a wave pattern that is one and a half wavelengths long. This indicates that the string is vibrating at a higher harmonic, resulting in a more complex wave pattern with more nodes and antinodes compared to a string vibrating at its fundamental frequency or a lower harmonic.

Image summary: The figure is an illustration of a string's shape at two different moments. The image depicts a wave pattern along the string. The wave appears to be oscillating, showing how the string's displacement varies over time.

2 Definitions

Definition 1: Constructive Interference: The superposition of waves resulting in an increased amplitude.

Definition 2: Destructive Interference: The superposition of waves resulting in a decreased amplitude.

This cancellation is called destructive interference. Midway between the nodes are the points of greatest amplitude, or the antinodes, marked A. At the antinodes the displacements of the two waves in red and blue are always identical, giving a large resultant displacement; this phenomenon is called constructive interference. We can see from the figure that the distance between successive nodes or between successive antinodes is one half-wavelength, or lambda/2 .

We can derive a wave function for the standing wave of figure 15.24 by adding the wave functions y 1(x, t) and y 2(x, t) for two waves with equal amplitude, period, and wavelength traveling in opposite directions. Here y 1(x, t) (the red curves in figure 15.24) represents an incoming, or incident, wave traveling to the left along the +-x-axis, arriving at the point x = 0 and being reflected; y 2(x, t) (the blue curves in figure 15.24) represents the reflected wave traveling to the right from x = 0. We noted in Section 15.6 that the wave reflected from a fixed end of a string is inverted, so we give a negative sign to one of the waves:

Math summary: The expression defines two wave functions. The first represents an incident wave traveling to the left, while the second represents a reflected wave traveling to the right.

The change in sign corresponds to a shift in phase of 180 degrees or pi radians. At x equals 0 the motion from the reflected wave is A times cosine omega t and the motion from the incident wave is negative A times cosine omega t, which we can also write as A times cosine of the quantity omega t plus pi. From Equation 15.27, the wave function for the standing wave is the sum of the individual wave functions:

Math summary: This expression calculates the sum of two wave functions to describe a standing wave. It takes two cosine functions with slightly different arguments, adds them together, and scales the result by an amplitude factor.

We can rewrite each of the cosine terms by using the identities for the cosine of the sum and difference of two angles: cosine of (a plus or minus b) equals cosine of a times cosine of b minus or plus sine of a times sine of b. Applying these and combining terms, we obtain the wave function for the standing wave:

Math summary: This expression describes a standing wave on a string with a fixed end, where the wave function is determined by the product of a sine function dependent on position and a sine function dependent on time. The amplitude of the standing wave is twice the amplitude of the original traveling waves.

The standing-wave amplitude A sub S W is twice the amplitude A of either of the original traveling waves: A sub S W equals 2 times A.

Equation (15.28) has two factors: a function of x and a function of t. The factor A sub sw is a wave traveling along a string, the wave shape stays in the same position, oscillating up and down as described by the sinomega t factor. This behavior is shown by the gold curves in figure 15.24. Each point in the string still undergoes simple harmonic motion, but all the points between any successive pair of nodes oscillate in phase. This is in contrast to the phase differences between oscillations of adjacent points that we see with a traveling wave.

We can use equation (15.28) to find the positions of the nodes; these are the points for which sine of k times x equals 0, so the displacement is always zero. This occurs when k times x equals 0, pi, 2 times pi, 3 times pi, and so on, or, using k equals 2 times pi divided by lambda,

Math summary: This expression identifies the positions of nodes, or points of zero displacement, along a standing wave on a string with a fixed end. It provides a sequence of locations representing these nodes, expressed both in terms of a constant k and in terms of the wavelength lambda.

In particular, there is a node at x = 0, as there should be, since this point is a fixed end of the string.

A standing wave, unlike a traveling wave, does not transfer energy from one end to the other. The two waves that form it would individually carry equal amounts of power in opposite directions. There is a local flow of energy from each node to the adjacent antinodes and back, but the average rate of energy transfer is zero at every point. If you use the wave function of equation (15.28) to evaluate the wave power given by equation (15.21), you will find that the average power is zero.

Problem-Solving Strategy 15.2 Standing Waves

identify the relevant concepts: Identify the target variables. Then determine whether the problem is purely kinematic (involving only such quantities as wave speed v, wavelength lambda , and frequency f) or whether dynamic properties of the medium (such as F and mu for transverse waves on a string) are also involved.

set up the problem using the following steps:

1. Sketch the shape of the standing wave at a particular instant. This will help you visualize the nodes (label them N) and antinodes (A). The distance between adjacent nodes (or antinodes) is lambda/2 ; the distance between a node and the adjacent antinode is lambda/4 .

2. Choose the equations you'll use. The wave function for the standing wave, like equation (15.28), is often useful. 3. You can determine the wave speed if you know lambda and f (or, equivalently, k equals 2 times pi divided by lambda and omega equals 2 times pi times f) or if you know the relevant properties of the medium (for a string, F and mu).

execute the solution: Solve for the target variables. Once you've found the wave function, you can find the displacement y at any point 10 and at any time t. You can find the velocity and acceleration of a particle in the medium by taking the first and second partial derivatives of y with respect to time.

evaluate your answer: Compare your numerical answers with your sketch. Check that the wave function satisfies the boundary conditions (for example, the displacement should be zero at a fixed end).

Example 15.6 Standing Waves on a Guitar String

A guitar string lies along the x-axis when in equilibrium. The end of the string at x = 0 (the bridge of the guitar) is fixed. A sinusoidal wave with amplitude A = 0.750 millimeters = 7.50 times 10 sup -4 m and frequency f = 440 Hz , corresponding to the red curves in figure 15.24, travels along the string in the -x -direction at 143 meters per second. It is reflected from the fixed end, and the superposition of the incident and reflected waves forms a standing wave. (a) Find the equation giving the displacement of a point on the string as a function of position and time. (b) Locate the nodes. (c) Find the amplitude of the standing wave and the maximum transverse velocity and acceleration.

identify and set up This is a kinematics problem (see Problem-Solving Strategy 15.1 in Section 15.3). The target variables are: in part (a), the wave function of the standing wave; in part (b), the locations of the nodes; and in part (c), the maximum displacement y, transverse velocity v sub y, and transverse acceleration a sub y. Since there is a fixed end at x equals 0, we can use Eqs. (15.28) and (15.29) to describe this standing wave. We'll need the relationships omega equals 2 times pi times f, v equals omega divided by k, and v equals lambda times f.

execute (a) The standing-wave amplitude is A sub S W equals 2 times A equals 1.50 times 10 to the power of negative 3 meters (twice the amplitude of either the incident or reflected wave). The angular frequency and wave number are

Math summary: This calculates the angular frequency by multiplying two pi radians by the frequency of 440 per second, resulting in 2760 radians per second. Then, it calculates the wave number by dividing the angular frequency of 2760 radians per second by the velocity of 143 meters per second, resulting in 19.3 radians per meter.

Equation (15.28) then gives

Math summary: This expression calculates the displacement of a standing wave. It takes position and time as inputs, then computes the displacement as a product of sine functions with given amplitude, wave number, and angular frequency.

(b) From Equation 15.29, the positions of the nodes are x equals 0, lambda divided by 2, lambda, 3 times lambda divided by 2, and so on. The wavelength is lambda equals v divided by f equals 143 meters per second divided by 440 Hertz equals 0.325 meters, so the nodes are at x equals 0, 0.163 meters, 0.325 meters, 0.488 meters, and so on.

(c) From the expression for y of x and t in part (a), the maximum displacement from equilibrium is A sub S.W. equals 1.50 times 10 to the power of negative 3 meters equals 1.50 millimeters. This occurs at the antinodes, which are midway between adjacent nodes (that is, at x equals 0.081 meters, 0.244 meters, 0.406 meters, and so on).

For a particle on the string at any point 10, the transverse (y-) velocity is

Math summary: This calculates the transverse velocity of a particle on a string. It multiplies the sine of a position-dependent term by the cosine of a time-dependent term, scaling the result by a constant factor to determine the velocity.

At an antinode, sine of 19.3 radians per meter, times x, equals plus or minus 1 and the transverse velocity varies between plus 4.15 meters per second and minus 4.15 meters per second. As is always the case in S.H.M., the maximum velocity occurs when the particle is passing through the equilibrium position y equals 0.

The transverse acceleration a sub y of x comma t is the second partial derivative of y of x comma t with respect to time. You can show that

Math summary: This expression calculates the transverse acceleration as the second partial derivative of a function with respect to time. It shows how acceleration changes over time.

At the antinodes, the transverse acceleration varies between positive 1.15 times 10 to the power of 4 meters per second squared and negative 1.15 times 10 to the power of 4 meters per second squared.

evaluate The maximum transverse velocity at an antinode is quite respectable (about 15 kilometers/h, or 9.3 mi/h). But the maximum transverse acceleration is tremendous, 1170 times the acceleration due to gravity! Guitar strings are actually fixed at both ends; we'll see the consequences of this in the next section.

keyconcept A sinusoidal standing wave is the superposition of two sinusoidal waves of the same amplitude and frequency traveling in opposite directions. The wave nodes, which are spaced one half-wavelength apart, are points where the displacement in the standing wave is always zero.

test your understanding of section 15.7 Suppose the frequency of the standing wave in Example 15.6 were doubled from 440 Hz to 880 Hz. Would all of the nodes for f = 440 Hz also be nodes for f = 880 Hz? If so, would there be additional nodes for f = 880 Hz? If not, which nodes are absent for f = 880 Hz?

Answer

Yes, yes Doubling the frequency makes the wavelength half as large. Hence the spacing between nodes (equal to lambda/2 ) is also half as large. There are nodes at all of the previous positions, but there

15.8 Normal Modes of a String

When we described standing waves on a string rigidly held at one end, as in figure 15.23, we made no assumptions about the length of the string or about what was happening at the other end. Let's now consider a string of a definite length L, rigidly held at both ends. Such strings are found in many musical instruments, including pianos, violins, and guitars.

When a guitar string is plucked, a wave is produced in the string; this wave is reflected and re-reflected from the ends of the string, making a standing wave. This standing wave on the string in turn produces a sound wave in the air, with a frequency determined by the properties of the string. This is what makes stringed instruments so useful in making music.

To understand a standing wave on a string fixed at both ends, we first note that the standing wave must have a node at both ends of the string. We saw in the preceding section that adjacent nodes are one half-wavelength ( lambda/2 ) apart, so the length of the string must be lambda/2 , or 2(lambda/2) , or 3(lambda/2) , or in general some integer number of half-wavelengths:

Math summary: This equation calculates the possible lengths of a string fixed at both ends that can support a standing wave. The length is determined by multiplying an integer number by half of the wavelength.

That is, if a string with length L is fixed at both ends, a standing wave can exist only if its wavelength satisfies equation (15.30).

Solving this equation for lambda and labeling the possible values of lambda as lambda sub n, we find

Math summary: This equation calculates the possible wavelengths for standing waves on a string that is fixed at both ends. It determines these wavelengths by dividing twice the length of the string by a positive integer, where the integer represents the mode number of the standing wave.

Waves can exist on the string if the wavelength is not equal to one of these values, but there cannot be a steady wave pattern with nodes and antinodes, and the total wave cannot be a standing wave. Equation (15.31) is illustrated by the standing waves shown in Figs. 15.23a, 15.23b, 15.23c, and 15.23d; these represent n = 1, 2, 3, and 4, respectively.

Corresponding to the series of possible standing-wave wavelengths lambda sub n is a series of possible standing-wave frequencies f sub n, each related to its corresponding wavelength by f sub n equals v divided by lambda sub n. The smallest frequency f sub 1 corresponds to the largest wavelength (the n equals 1 case), lambda sub 1 equals 2 times L:

Math summary: This expression calculates the fundamental frequency of a vibrating string fixed at both ends. It divides the wave speed by twice the length of the string to determine the fundamental frequency.

This is called the fundamental frequency. The other standing-wave frequencies are f sub 2 equals 2 times v divided by 2 times L, f sub 3 equals 3 times v divided by 2 times L, and so on. These are all integer multiples of f sub 1, such as 2 times f sub 1, 3 times f sub 1, 4 times f sub 1, and so on. We can express all the frequencies as

Math summary: This calculates the standing wave frequencies for a string fixed at both ends. The standing wave frequencies are integer multiples of the fundamental frequency, where the fundamental frequency is the ratio of the wave's speed to twice the string's length.

These frequencies are called harmonics, and the series is called a harmonic series. Musicians sometimes call f sub 2, f sub 3, and so on overtones; f sub 2 is the second harmonic or the first overtone, f sub 3 is the third harmonic or the second overtone, and so on. The first harmonic is the same as the fundamental frequency (Fig. 15.25).

Figure 15.25 summary: The figure is a photograph. It depicts a violin being played with a bow. The violin strings vibrate at harmonic frequencies. These vibrations generate sound waves at the same frequencies in the surrounding air. The player uses the bow to excite the strings, causing them to oscillate and produce musical notes.

For a string with fixed ends at x equals 0 and x equals L, the wave function y of x comma t of the nth standing wave is given by Equation fifteen point twenty eight, which satisfies the condition that there is a node at x equals 0, with omega equals omega sub n equals 2 times pi times f sub n and k equals k sub n equals 2 times pi divided by lambda sub n:

Math summary: This equation calculates the wave function of the nth standing wave on a string. It takes the amplitude of the standing wave and multiplies it by the sine of the wave number times position and the sine of the angular frequency times time.

You can confirm that this wave function has nodes at both x = 0 and x = L.

Definition

Normal Mode: A specific pattern of vibration in a system where all parts move sinusoidally with the same frequency.

A normal mode of an oscillating system is a motion in which all particles of the system move sinusoidally with the same frequency. For a system made up of a string of length L fixed at both ends, each of the frequencies given by equation (15.33) corresponds to a possible normal-mode pattern. Figure 15.26 shows the first four normal-mode patterns and their associated frequencies and wavelengths; these correspond to equation (15.34) with n = 1, 2, 3, and 4. By contrast, a harmonic oscillator, which has only one oscillating particle, has only one normal mode and one characteristic frequency. The string fixed at both ends has infinitely many normal modes (n = 1, 2, 3, ... ) because it is made up of a very large (effectively infinite) number of particles. More complicated oscillating systems also have infinite numbers of normal modes, though with more complex normal-mode patterns (Fig. 15.27, next page).

Figure 15.26 summary: The figure is a depiction of standing waves. The figure illustrates the fundamental frequency and the second harmonic. The fundamental frequency shows a single segment, while the second harmonic displays two segments. The higher the harmonic, the more segments are present within the standing wave.

Figure 15.27 summary: This is an illustration of a cross section of the sun's interior. The figure depicts the internal structure of the sun, revealing different layers or regions. The core is visibly distinct from the outer layers. The illustration provides a visualization of the sun's composition and internal dynamics.

Image summary: The image is a representation of a standing wave, specifically depicting the fourth harmonic. The wave is fixed at both ends, showing several nodes and antinodes. The length of the string corresponds to a specific number of half-wavelengths. The standing wave illustrates a higher harmonic mode, implying a shorter wavelength and consequently a higher frequency compared to the fundamental mode.

Figure 15.27 Astronomers have discovered that the sun oscillates in several different normal modes. This computer simulation shows one such mode.

Complex Standing Waves

If we could displace a string so that its shape is the same as one of the normal-mode patterns and then release it, it would vibrate with the frequency of that mode. Such a vibrating string would displace the surrounding air with the same frequency, producing a traveling sinusoidal sound wave that your ears would perceive as a pure tone. But when a string is struck (as in a piano) or plucked (as is done to guitar strings), the shape of the displaced string is not one of the patterns in figure 15.26. The motion is therefore a combination or superposition of many normal modes.

Several simple harmonic motions of different frequencies are present simultaneously, and the displacement of any point on the string is the superposition of the displacements associated with the individual modes. The sound produced by the vibrating string is likewise a superposition of traveling sinusoidal sound waves, which you perceive as a rich, complex tone with the fundamental frequency f sub 1 . The standing wave on the string and the traveling sound wave in the air have similar harmonic content (the extent to which frequencies higher than the fundamental are present). The harmonic content depends on how the string is initially set into motion.

If you pluck the strings of an acoustic guitar in the normal location over the sound hole, the sound that you hear has a different harmonic content than if you pluck the strings next to the fixed end on the guitar body.

It is possible to represent every possible motion of the string as some superposition of normal-mode motions. Finding this representation for a given vibration pattern is called harmonic analysis. The sum of sinusoidal functions that represents a complex wave is called a Fourier series. Figure 15.28 shows how a standing wave that is produced by plucking a guitar string of length 50 at a point L/4 from one end can be represented as a combination of sinusoidal functions.

Figure 15.28 summary: The figure contains two line graphs. The first line graph displays three sinusoidal functions with different amplitudes and frequencies. The second line graph shows the approximation of a function using the sum of the three sinusoidal functions from the first graph. The approximation gets better as more terms are added, and the sum of the first three terms provides a reasonable, but not perfect, approximation.

Standing Waves and String Instruments

From equation (15.32), the fundamental frequency of a vibrating string is f sub 1 equals v divided by 2 times L. The speed v of waves on the string is determined by equation (15.14), v equals square root of F divided by mu. Combining these equations, we find

Math summary: This equation calculates the fundamental frequency of a vibrating string. It shows that the fundamental frequency is determined by the length of the string, the tension in the string, and a constant related to the string's mass per unit length.

This is also the fundamental frequency of the sound wave created in the surrounding air by the vibrating string. The inverse dependence of frequency on length L is illustrated by the long strings of the bass (low-frequency) section of the piano or the bass viol compared with the shorter strings of the treble section of the piano or the violin (Fig. 15.29). The pitch of a violin or guitar is usually varied by pressing a string against the fingerboard with the fingers to change the length L of the vibrating portion of the string. Increasing the tension F increases the wave speed v and thus increases the frequency (and the pitch). All string instruments are “tuned” to the correct frequencies by varying the tension; you tighten the string to raise the pitch.

Figure 15.29 summary: The figure is a comparative illustration. It shows the range of notes, from bass to treble, that can be played by a concert grand piano, a bass viol, a cello, a viola, and a violin. The piano has the widest range, encompassing the ranges of all the other instruments. The bass viol has the lowest range, while the violin has the highest range among the string instruments. Longer strings produce bass notes and shorter strings produce treble notes.

Finally, increasing the mass per unit length mu decreases the wave speed and thus the frequency. The lower notes on a steel guitar are produced by thicker strings, and one reason for winding the bass strings of a piano with wire is to obtain the desired low frequency from a relatively short string.

Wind instruments such as saxophones and trombones also have normal modes. As for stringed instruments, the frequencies of these normal modes determine the pitch of the musical tones that these instruments produce. We'll discuss these instruments and many other aspects of sound in Chapter 16.

Example 15.7 A Giant Bass Viol

In an attempt to get your name in Guinness World Records, you build a bass viol with strings of length 5.00 m between fixed points. One string, with linear mass density 40.0 g/m, is tuned to a 20.0 Hz fundamental frequency (the lowest frequency that the human ear can hear). Calculate (a) the tension of this string, (b) the frequency and wavelength on the string of the second harmonic, and (c) the frequency and wavelength on the string of the second overtone.

identify and set up In part (a) the target variable is the string tension F; we'll use equation (15.35), which relates F to the known values f sub 1 equals 20.0 Hertz, L equals 5.00 meters, and mu equals 40.0 grams per meter squared. In parts (b) and (c) the target variables are the frequency and wavelength of a given harmonic and a given overtone. We determine these from the given length of the string and the fundamental frequency, using Eqs. (15.31) and (15.33).

With Variation Problems

execute (a) We solve equation (15.35) for F:

Math summary: This calculates the tension using the linear density, length, and fundamental frequency. It multiplies four times the linear density by the square of the length and the square of the fundamental frequency to produce the tension, which is equal to sixteen hundred Newtons or three hundred sixty pounds.

(b) From Eqs. (15.33) and (15.31), the frequency and wavelength of the second harmonic (n = 2) are

Math summary: This calculates the frequency and wavelength of the second harmonic. It doubles the fundamental frequency of 20.0 Hertz to obtain a second harmonic frequency of 40.0 Hertz, and it calculates the wavelength as two times the length of 5.00 meters divided by two, resulting in a wavelength of 5.00 meters.

(c) The second overtone is the “second tone over” (above) the fundamental—that is, n = 3. Its frequency and wavelength are

Math summary: First, the frequency of the second overtone is calculated by multiplying the fundamental frequency by three, resulting in 60.0 Hertz. Then, the wavelength is determined by multiplying twice the length by one third, which equals 3.33 meters.

evaluate The string tension in a real bass viol is typically a few hundred newtons; the tension in part (a) is a bit higher than that. The wavelengths in parts (b) and (c) are equal to the length of the string and two-thirds the length of the string, respectively, which agrees with the drawings of standing waves in figure 15.26.

keyconcept A string with its ends fixed vibrates in a type of standing wave called a normal mode when all of its particles move sinusoidally with the same frequency. The frequencies of oscillation of the normal modes are integer multiples of a minimum normal-mode frequency, called the fundamental frequency.

Example 15.8 From Waves on a String to Sound Waves in Air

What are the frequency and wavelength of the sound waves produced in the air when the string in Example 15.7 is vibrating at its fundamental frequency? The speed of sound in air at 20 sup circle C is 344 meters per second.

identify and set up Our target variables are the frequency and wavelength for the sound wave produced by the bass viol string. The frequency of the sound wave is the same as the fundamental frequency f sub 1 of the standing wave, because the string forces the surrounding air to vibrate at the same frequency. The wavelength of the sound wave is lambda sub 1 sound equals v sub sound divided by f sub 1.

execute We have f equals f sub 1 equals 20.0 Hertz, so

Math summary: This expression calculates the wavelength of a sound wave. It divides the speed of sound by the frequency to obtain the wavelength, resulting in a wavelength of 17.2 meters.

evaluate In Example 15.7, the wavelength of the fundamental on the string was lambda sub 1 open parenthesis string close parenthesis equals 2 times L equals 2 times open parenthesis 5.00 meters close parenthesis equals 10.0 meters. Here lambda sub 1 open parenthesis sound close parenthesis equals 17.2 meters is greater than that by the factor of 17.2 divided by 10.0 equals 1.72. This is as it should be: Because the frequencies of the sound wave and the standing wave are equal, lambda equals v divided by f says that the wavelengths in air and on the string are in the same ratio as the corresponding wave speeds; here v sub sound equals 344 meters per second is greater than v sub string equals open parenthesis 10.0 meters close parenthesis times open parenthesis 20.0 Hertz close parenthesis equals 200 meters per second by just the factor 1.72.

keyconcept A string vibrating at a certain frequency produces sound waves of the same frequency in the surrounding air. The standing wave on the string and the sound wave will have different wavelengths, however, if the speed of waves on the string does not equal the speed of sound.

test your understanding of section 15.8 While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. Which normal modes cannot be present on the string while you are touching it in this way?

Answer

In n = 1, 3, 5, . When you touch the string at its center, you are producing a node at the center. Hence only standing waves with a node at x = L/2 are allowed. From Figure 15.26 you can see that the normal modes n = 1, 3, 5, cannot be present.

Waves and their properties: A wave is any disturbance that propagates from one region to another. A mechanical wave travels within some material called the medium. The wave speed v depends on the type of wave and the properties of the medium.

In a periodic wave, the motion of each point of the medium is periodic with frequency f and period T. The wavelength lambda is the distance over which the wave pattern repeats, and the amplitude A is the maximum displacement of a particle in the medium. The product of lambda and f equals the wave speed. A sinusoidal wave is a special periodic wave in which each point moves in simple harmonic motion. (See Example 15.1.)

Math summary: This expression calculates the displacement of a point on a wave. It takes the cosine of a phase that depends on position, time, wave speed, and angular frequency, then scales the result by the amplitude to determine the displacement.

Wave functions and wave dynamics: The wave function y(x, t) describes the displacements of individual particles in the medium. Equations (15.3), (15.4), and (15.7) give the wave equation for a sinusoidal wave traveling in the +x-direction. If the wave is moving in the -x-direction, the minus signs in the cosine functions are replaced by plus signs. (See Example 15.2.)

Math summary: This expression calculates the displacement of a particle in a medium as a function of position and time. It takes position and time as inputs, scales them by wavelength and period respectively, and then computes a cosine function to determine the displacement.

Image summary: The image is a line graph. The graph shows a wave oscillating along the x-axis. The wave has a maximum amplitude of A and a minimum amplitude of negative A. The distance between two consecutive peaks or troughs is labeled as the wavelength lambda. The wave oscillates symmetrically around the x-axis, with the peaks and troughs being equidistant from the x-axis.

Math summary: This expression represents a sinusoidal wave function. It calculates the displacement at a given position and time using the cosine of the position and time, scaled by constants for amplitude, wave number, and angular frequency.

The wave function obeys a partial differential equation called the wave equation, equation (15.12).

Math summary: This expression defines the relationship between wave number, wavelength, angular frequency, frequency, and velocity. It shows that wave number equals two pi divided by wavelength, and angular frequency equals two pi times frequency, which also equals velocity times wave number.

The speed of transverse waves on a string depends on the tension F and mass per unit length mu . (See Example 15.3.)

Math summary: This equation relates the second spatial derivative of a wave function to its second time derivative. It states that the spatial curvature of the wave at any point is proportional to its acceleration in time at that point, scaled by the inverse square of the wave's velocity.

Image summary: The image is a line plot. The plot shows a sinusoidal wave, depicting the relationship between a variable y and time t. The wave oscillates symmetrically above and below the horizontal axis. The maximum displacement from the horizontal axis, labeled as A, represents the amplitude of the wave. The horizontal distance between two consecutive peaks, labeled as T, represents the period of the wave. The wave exhibits a periodic behavior, repeating its pattern over time. The wave oscillates between positive and negative values, indicating a fluctuating quantity.

Math summary: This expression calculates the speed of waves on a string. It takes the square root of the tension divided by the mass per unit length to determine the wave speed.

Wave power: Wave motion conveys energy from one region to another. For a sinusoidal mechanical wave, the average power P sub av is proportional to the square of the wave amplitude and the square of the frequency. For waves that spread out in three dimensions, the wave intensity I is inversely proportional to the square of the distance from the source. (See Examples 15.4 and 15.5.)

Math summary: This expression calculates the average power of a sinusoidal wave. It takes the square root of the product of mass per unit length and tension, multiplies it by one half, and then multiplies by the square of both the angular frequency and the amplitude squared.

(average power, sinusoidal wave)

Wave power versus time t at coordinate x = 0

Math summary: This equation calculates the ratio of two intensities. It states that the ratio of the first intensity to the second intensity is equal to the ratio of the second radius squared to the first radius squared.

(inverse-square law for intensity)

Image summary: The figure is a line plot. The plot shows the relationship between power and time. The power varies periodically with time, oscillating between zero and a maximum value. The average power is half of the maximum power.

Wave superposition: A wave reflects when it reaches a boundary of its medium. At any point where two or more waves overlap, the total displacement is the sum of the displacements of the individual waves (principle of superposition).

Math summary: This expression calculates the superposition of waves. It determines the total displacement at a point by summing the displacements of individual waves.

(principle of superposition)

Image summary: The image displays a series of snapshots depicting the superposition of two wave pulses. The figure illustrates wave pulses moving towards each other, interacting, and then continuing to propagate in their original directions. The image shows the wave pulses before, during, and after they overlap. The resulting amplitude during the interaction is determined by the principle of superposition. The figure suggests that after the interaction, the wave pulses maintain their original shapes and directions, demonstrating the wave nature of these disturbances.

Standing waves on a string: When a sinusoidal wave is reflected from a fixed or free end of a stretched string, the incident and reflected waves combine to form a standing sinusoidal wave with nodes and antinodes. Adjacent nodes are spaced a distance lambda/2 apart, as are adjacent antinodes. (See Example 15.6.)

Math summary: This expression calculates the displacement of a standing wave on a string. It multiplies the amplitude of the standing wave by the sine of the wave number times position, and then multiplies that result by the sine of the angular frequency times time.

(standing wave on a string, fixed end at x = 0)

Math summary: This expression calculates the frequencies of standing waves on a string fixed at both ends. It determines these frequencies by multiplying the mode number, which is an integer, by the fundamental frequency.

When both ends of a string with length L are held fixed, standing waves can occur only when L is an integer multiple of lambda/2 . Each frequency with its associated vibration patterns called a normal mode. (See Examples 15.7 and 15.8.)

Image summary: The image is a diagram that depicts standing waves on a string fixed at both ends. The diagram illustrates two different modes of vibration. The first mode shows a single antinode in the middle, with the length of the string equal to one-half of the wavelength. The second mode shows two antinodes, with the length of the string equal to a full wavelength. The diagram shows that different modes of vibration can exist on a string fixed at both ends, each corresponding to a different wavelength and number of antinodes. The wavelength is inversely proportional to the number of antinodes.

Math summary: This formula calculates the fundamental frequency of a vibrating string. It takes the square root of the tension divided by the linear density, and then divides that result by twice the length of the string.

(string fixed at both ends)

Image summary: The image shows a depiction of standing waves. The image illustrates two standing wave patterns within a fixed length. The standing wave in the second depiction has a higher frequency and shorter wavelength compared to the standing wave in the first depiction. The number of antinodes and nodes is greater in the second standing wave pattern.

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