Chapter 18. Superposition and Standing Waves

Transcription

1 Chapter 18 Superposition and Standing Waes C HP T E R O UTLI N E 181 Superposition and Interference 182 Standing Waes 183 Standing Waes in a String Fixed at Both Ends 184 Resonance 185 Standing Waes in ir Columns 186 Standing Waes in Rods and Membranes 187 Beats: Interference in Time 188 Nonsinusoidal Wae Patterns Guitarist Carlos Santana takes adantage of standing waes on strings. He changes to a higher note on the guitar by pushing the strings against the frets on the fingerboard, shortening the lengths of the portions of the strings that ibrate. (Bettmann/Corbis)

2 In the preious two chapters, we introduced the wae model. We hae seen that waes are ery different from particles. particle is of zero size, while a wae has a characteristic size the waelength. nother important difference between waes and particles is that we can explore the possibility of two or more waes combining at one point in the same medium. We can combine particles to form extended objects, but the particles must be at different locations. In contrast, two waes can both be present at the same location, and the ramifications of this possibility are explored in this chapter. When waes are combined, only certain allowed frequencies can exist on systems with boundary conditions the frequencies are quantized. Quantization is a notion that is at the heart of quantum mechanics, a subject that we introduce formally in Chapter 40. There we show that waes under boundary conditions explain many of the quantum phenomena. For our present purposes in this chapter, quantization enables us to understand the behaior of the wide array of musical instruments that are based on strings and air columns. We also consider the combination of waes haing different frequencies and waelengths. When two sound waes haing nearly the same frequency interfere, we hear ariations in the loudness called beats. The beat frequency corresponds to the rate of alternation between constructie and destructie interference. Finally, we discuss how any nonsinusoidal periodic wae can be described as a sum of sine and cosine functions Superposition and Interference Many interesting wae phenomena in nature cannot be described by a single traeling wae. Instead, one must analyze complex waes in terms of a combination of traeling waes. To analyze such wae combinations, one can make use of the superposition principle: Superposition principle If two or more traeling waes are moing through a medium, the resultant alue of the wae function at any point is the algebraic sum of the alues of the wae functions of the indiidual waes. Waes that obey this principle are called linear waes. In the case of mechanical waes, linear waes are generally characterized by haing amplitudes much smaller than their waelengths. Waes that iolate the superposition principle are called nonlinear waes and are often characterized by large amplitudes. In this book, we deal only with linear waes. One consequence of the superposition principle is that two traeling waes can pass through each other without being destroyed or een altered. For instance, 544

3 SECTION 18.1 Superposition and Interference 545 (a) y 1 y 2 (b) y 1 + y 2 (c) (d) y 1 + y 2 y 2 y 1 (e) Education Deelopment Center, Newton, M ctie Figure 18.1 (a d) Two pulses traeling on a stretched string in opposite directions pass through each other. When the pulses oerlap, as shown in (b) and (c), the net displacement of the string equals the sum of the displacements produced by each pulse. Because each pulse produces positie displacements of the string, we refer to their superposition as constructie interference. (e) Photograph of the superposition of two equal, symmetric pulses traeling in opposite directions on a stretched spring. t the ctie Figures link at you can choose the amplitude and orientation of each of the pulses and study the interference between them as they pass each other. when two pebbles are thrown into a pond and hit the surface at different places, the expanding circular surface waes do not destroy each other but rather pass through each other. The complex pattern that is obsered can be iewed as two independent sets of expanding circles. Likewise, when sound waes from two sources moe through air, they pass through each other. Figure 18.1 is a pictorial representation of the superposition of two pulses. The wae function for the pulse moing to the right is y 1, and the wae function for the pulse moing to the left is y 2. The pulses hae the same speed but different shapes, and the displacement of the elements of the medium is in the positie y direction for both pulses. When the waes begin to oerlap (Fig. 18.1b), the wae function for the resulting complex wae is gien by y 1 y 2. When the crests of the pulses coincide (Fig. 18.1c), the resulting wae gien by y 1 y 2 has a larger amplitude than that of the indiidual pulses. The two pulses finally separate and continue moing in their original directions (Fig. 18.1d). Note that the pulse shapes remain unchanged after the interaction, as if the two pulses had neer met! The combination of separate waes in the same region of space to produce a resultant wae is called interference. For the two pulses shown in Figure 18.1, the displacement of the elements of the medium is in the positie y direction for both pulses, and the resultant pulse (created when the indiidual pulses oerlap) exhibits an amplitude greater than that of either indiidual pulse. Because the displacements caused by the two pulses are in the same direction, we refer to their superposition as constructie interference. Now consider two pulses traeling in opposite directions on a taut string where one pulse is inerted relatie to the other, as illustrated in Figure In this case, when the pulses begin to oerlap, the resultant pulse is gien by y 1 y 2, but the alues of the function y 2 are negatie. gain, the two pulses pass through each other; howeer, because the displacements caused by the two pulses are in opposite directions, we refer to their superposition as destructie interference. PITFLL PREVENTION 18.1 Do Waes Really Interfere? In popular usage, the term interfere implies that an agent affects a situation in some way so as to preclude something from happening. For example, in merican football, pass interference means that a defending player has affected the receier so that he is unable to catch the ball. This is ery different from its use in physics, where waes pass through each other and interfere, but do not affect each other in any way. In physics, interference is similar to the notion of combination as described in this chapter. Constructie interference Destructie interference

4 546 CHPTER 18 Superposition and Standing Waes (a) y 1 y 2 (b) y 1 y 2 (c) y 1 + y 2 t the ctie Figures link at you can choose the amplitude and orientation of each of the pulses and watch the interference as they pass each other. (d) (e) y 2 y 2 y 1 y 1 ctie Figure 18.2 (a e) Two pulses traeling in opposite directions and haing displacements that are inerted relatie to each other. When the two oerlap in (c), their displacements partially cancel each other. (f ) Photograph of the superposition of two symmetric pulses traeling in opposite directions, where one is inerted relatie to the other. (f) Education Deelopment Center, Newton, M Quick Quiz 18.1 Two pulses are traeling toward each other, each at 10 cm/s on a long string, as shown in Figure Sketch the shape of the string at t 0.6 s. 1 cm Figure 18.3 (Quick Quiz 18.1) The pulses on this string are traeling at 10 cm/s. Quick Quiz 18.2 Two pulses moe in opposite directions on a string and are identical in shape except that one has positie displacements of the elements of the string and the other has negatie displacements. t the moment that the two pulses completely oerlap on the string, (a) the energy associated with the pulses has disappeared (b) the string is not moing (c) the string forms a straight line (d) the pulses hae anished and will not reappear.

5 SECTION 18.1 Superposition and Interference 547 Superposition of Sinusoidal Waes Let us now apply the principle of superposition to two sinusoidal waes traeling in the same direction in a linear medium. If the two waes are traeling to the right and hae the same frequency, waelength, and amplitude but differ in phase, we can express their indiidual wae functions as y 1 sin(kx t) y 2 sin(kx t ) where, as usual, k 2/, 2f, and is the phase constant, which we discussed in Section Hence, the resultant wae function y is y y 1 y 2 [sin(kx t) sin(kx t )] To simplify this expression, we use the trigonometric identity sin a sin b 2 cos a b 2 sin a b 2 If we let a kx t and b kx t, we find that the resultant wae function y reduces to y 2 cos 2 sin kx t 2 Resultant of two traeling sinusoidal waes This result has seeral important features. The resultant wae function y also is sinusoidal and has the same frequency and waelength as the indiidual waes because the sine function incorporates the same alues of k and that appear in the original wae functions. The amplitude of the resultant wae is 2 cos(/2), and its phase is /2. If the phase constant equals 0, then cos (/2) cos 0 1, and the amplitude of the resultant wae is 2 twice the amplitude of either indiidual wae. In this case the waes are said to be eerywhere in phase and thus interfere constructiely. That is, the crests and troughs of the indiidual waes y 1 and y 2 occur at the same positions and combine to form the red cure y of amplitude 2 shown in Figure 18.4a. Because the y y y 1 and y 2 are identical φ = 0 x (a) y y 1 y 2 y x φ = 180 y φ = 60 y (b) (c) y 1 y 2 x ctie Figure 18.4 The superposition of two identical waes y 1 and y 2 (blue and green) to yield a resultant wae (red). (a) When y 1 and y 2 are in phase, the result is constructie interference. (b) When y 1 and y 2 are rad out of phase, the result is destructie interference. (c) When the phase angle has a alue other than 0 or rad, the resultant wae y falls somewhere between the extremes shown in (a) and (b). t the ctie Figures link at you can change the phase relationship between the waes and obsere the wae representing the superposition.

6 548 CHPTER 18 Superposition and Standing Waes indiidual waes are in phase, they are indistinguishable in Figure 18.4a, in which they appear as a single blue cure. In general, constructie interference occurs when cos(/2) 1. This is true, for example, when 0, 2, 4,... rad that is, when is an een multiple of. When is equal to rad or to any odd multiple of, then cos(/2) cos(/2) 0, and the crests of one wae occur at the same positions as the troughs of the second wae (Fig. 18.4b). Thus, the resultant wae has zero amplitude eerywhere, as a consequence of destructie interference. Finally, when the phase constant has an arbitrary alue other than 0 or an integer multiple of rad (Fig. 18.4c), the resultant wae has an amplitude whose alue is somewhere between 0 and 2. Interference of Sound Waes One simple deice for demonstrating interference of sound waes is illustrated in Figure Sound from a loudspeaker S is sent into a tube at point P, where there is a T-shaped junction. Half of the sound energy traels in one direction, and half traels in the opposite direction. Thus, the sound waes that reach the receier R can trael along either of the two paths. The distance along any path from speaker to receier is called the path length r. The lower path length r 1 is fixed, but the upper path length r 2 can be aried by sliding the U-shaped tube, which is similar to that on a slide trombone. When the difference in the path lengths r r 2 r 1 is either zero or some integer multiple of the waelength (that is r n, where n 0, 1, 2, 3,...), the two waes reaching the receier at any instant are in phase and interfere constructiely, as shown in Figure 18.4a. For this case, a maximum in the sound intensity is detected at the receier. If the path length r 2 is adjusted such that the path difference r /2, 3/2,..., n/2 (for n odd), the two waes are exactly rad, or 180, out of phase at the receier and hence cancel each other. In this case of destructie interference, no sound is detected at the receier. This simple experiment demonstrates that a phase difference may arise between two waes generated by the same source when they trael along paths of unequal lengths. This important phenomenon will be indispensable in our inestigation of the interference of light waes in Chapter 37. It is often useful to express the path difference in terms of the phase angle between the two waes. Because a path difference of one waelength corresponds to a phase angle of 2 rad, we obtain the ratio /2 r/ or Relationship between path difference and phase angle r 2 (18.1) Using the notion of path difference, we can express our conditions for constructie and destructie interference in a different way. If the path difference is any een multiple of /2, then the phase angle 2n, where n 0, 1, 2, 3,..., and the interference is constructie. For path differences of odd multiples of /2, (2n 1), where n 0, 1, 2, 3,..., and the interference is destructie. Thus, we hae the conditions S r 2 Speaker P r 1 R Receier Figure 18.5 n acoustical system for demonstrating interference of sound waes. sound wae from the speaker (S) propagates into the tube and splits into two parts at point P. The two waes, which combine at the opposite side, are detected at the receier (R). The upper path length r 2 can be aried by sliding the upper section.

7 SECTION Standing Waes 549 r (2n) 2 for constructie interference and (18.2) r (2n 1) 2 for destructie interference This discussion enables us to understand why the speaker wires in a stereo system should be connected properly. When connected the wrong way that is, when the positie (or red) wire is connected to the negatie (or black) terminal on one of the speakers and the other is correctly wired the speakers are said to be out of phase one speaker cone moes outward while the other moes inward. s a consequence, the sound wae coming from one speaker destructiely interferes with the wae coming from the other along a line midway between the two, a rarefaction region due to one speaker is superposed on a compression region from the other speaker. lthough the two sounds probably do not completely cancel each other (because the left and right stereo signals are usually not identical), a substantial loss of sound quality occurs at points along this line. Example 18.1 Two Speakers Drien by the Same Source pair of speakers placed 3.00 m apart are drien by the same oscillator (Fig. 18.6). listener is originally at point O, which is located 8.00 m from the center of the line connecting the two speakers. The listener then walks to point P, which is a perpendicular distance m from O, before reaching the first minimum in sound intensity. What is the frequency of the oscillator? Solution To find the frequency, we must know the waelength of the sound coming from the speakers. With this information, combined with our knowledge of the speed of sound, we can calculate the frequency. The waelength can be determined from the interference information gien. The first minimum occurs when the two waes reaching the listener at point P are 180 out of phase in other words, when their path difference r equals /2. To calculate the path difference, we must first find the path lengths r 1 and r m 1.15 m r 1 r m P O 1.85 m 8.00 m Figure 18.6 (Example 18.1) Two speakers emit sound waes to a listener at P. Figure 18.6 shows the physical arrangement of the speakers, along with two shaded right triangles that can be drawn on the basis of the lengths described in the problem. From these triangles, we find that the path lengths are and r 1 (8.00 m) 2 (1.15 m) m r 2 (8.00 m) 2 (1.85 m) m Hence, the path difference is r 2 r m. Because we require that this path difference be equal to /2 for the first minimum, we find that 0.26 m. To obtain the oscillator frequency, we use Equation 16.12, f, where is the speed of sound in air, 343 m/s: f 343 m/s 0.26 m 1.3 khz What If? What if the speakers were connected out of phase? What happens at point P in Figure 18.6? nswer In this situation, the path difference of /2 combines with a phase difference of /2 due to the incorrect wiring to gie a full phase difference of. s a result, the waes are in phase and there is a maximum intensity at point P Standing Waes The sound waes from the speakers in Example 18.1 leae the speakers in the forward direction, and we considered interference at a point in front of the speakers. Suppose that we turn the speakers so that they face each other and then hae them emit sound of the same frequency and amplitude. In this situation, two identical waes trael in

8 550 CHPTER 18 Superposition and Standing Waes opposite directions in the same medium, as in Figure These waes combine in accordance with the superposition principle. We can analyze such a situation by considering wae functions for two transerse sinusoidal waes haing the same amplitude, frequency, and waelength but traeling in opposite directions in the same medium: y 1 sin(kx t) y 2 sin(kx t) Figure 18.7 Two speakers emit sound waes toward each other. When they oerlap, identical waes traeling in opposite directions will combine to form standing waes. where y 1 represents a wae traeling in the x direction and y 2 represents one traeling in the x direction. dding these two functions gies the resultant wae function y: y y 1 y 2 sin(kx t) sin(kx t) When we use the trigonometric identity sin(a b) sin(a) cos(b) cos(a) sin(b), this expression reduces to y (2 sin kx) cos t (18.3) Equation 18.3 represents the wae function of a standing wae. standing wae, such as the one shown in Figure 18.8, is an oscillation pattern with a stationary outline that results from the superposition of two identical waes traeling in opposite directions. Notice that Equation 18.3 does not contain a function of kx t. Thus, it is not an expression for a traeling wae. If we obsere a standing wae, we hae no sense of motion in the direction of propagation of either of the original waes. If we compare this equation with Equation 15.6, we see that Equation 18.3 describes a special kind of simple harmonic motion. Eery element of the medium oscillates in simple harmonic motion with the same frequency (according to the cos t factor in the equation). Howeer, the amplitude of the simple harmonic motion of a gien element (gien by the factor 2 sin kx, the coefficient of the cosine function) depends on the location x of the element in the medium. The maximum amplitude of an element of the medium has a minimum alue of zero when x satisfies the condition sin kx 0, that is, when PITFLL PREVENTION 18.2 Three Types of mplitude We need to distinguish carefully here between the amplitude of the indiidual waes, which is, and the amplitude of the simple harmonic motion of the elements of the medium, which is 2 sin kx. gien element in a standing wae ibrates within the constraints of the enelope function 2 sin kx, where x is that element s position in the medium. This is in contrast to traeling sinusoidal waes, in which all elements oscillate with the same amplitude and the same frequency, and the amplitude of the wae is the same as the amplitude of the simple harmonic motion of the elements. Furthermore, we can identify the amplitude of the standing wae as 2. Because k 2/, these alues for kx gie x kx, 2, 3,... These points of zero amplitude are called nodes Richard Megna/Fundamental Photographs 2,, 3 2, ntinode n 2 n 0, 1, 2, 3, Node ntinode Node 2 sin kx Figure 18.8 Multiflash photograph of a standing wae on a string. The time behaior of the ertical displacement from equilibrium of an indiidual element of the string is gien by cos t. That is, each element ibrates at an angular frequency. The amplitude of the ertical oscillation of any elements of the string depends on the horizontal position of the element. Each element ibrates within the confines of the enelope function 2 sin kx. (18.4)

9 SECTION Standing Waes 551 The element with the greatest possible displacement from equilibrium has an amplitude of 2, and we define this as the amplitude of the standing wae. The positions in the medium at which this maximum displacement occurs are called antinodes. The antinodes are located at positions for which the coordinate x satisfies the condition sin kx 1, that is, when kx 2, 3 2, 5 2, Thus, the positions of the antinodes are gien by x 4, 3 4, 5 4, n 4 n 1, 3, 5, (18.5) Position of antinodes In examining Equations 18.4 and 18.5, we note the following important features of the locations of nodes and antinodes: The distance between adjacent antinodes is equal to /2. The distance between adjacent nodes is equal to /2. The distance between a node and an adjacent antinode is /4. Wae patterns of the elements of the medium produced at arious times by two waes traeling in opposite directions are shown in Figure The blue and green cures are the wae patterns for the indiidual traeling waes, and the red cures are the wae patterns for the resultant standing wae. t t 0 (Fig. 18.9a), the two traeling waes are in phase, giing a wae pattern in which each element of the medium is experiencing its maximum displacement from equilibrium. One quarter of a period later, at t T/4 (Fig. 18.9b), the traeling waes hae moed one quarter of a waelength (one to the right and the other to the left). t this time, the traeling waes are out of phase, and each element of the medium is passing through the equilibrium position in its simple harmonic motion. The result is zero displacement for elements at all alues of x that is, the wae pattern is a straight line. t t T/2 (Fig. 18.9c), the traeling waes are again in phase, producing a wae pattern that is inerted relatie to the t 0 pattern. In the standing wae, the elements of the medium alternate in time between the extremes shown in Figure 18.9a and c. y 1 y 1 y 1 y 2 y 2 y 2 y N N N N N y y N N N N N (a) t = 0 (b) t = T/4 (c) t = T/2 ctie Figure 18.9 Standing-wae patterns produced at arious times by two waes of equal amplitude traeling in opposite directions. For the resultant wae y, the nodes (N) are points of zero displacement, and the antinodes () are points of maximum displacement. t the ctie Figures link at you can choose the waelength of the waes and see the standing wae that results.

10 552 CHPTER 18 Superposition and Standing Waes Quick Quiz 18.3 Consider a standing wae on a string as shown in Figure Define the elocity of elements of the string as positie if they are moing upward in the figure. t the moment the string has the shape shown by the red cure in Figure 18.9a, the instantaneous elocity of elements along the string (a) is zero for all elements (b) is positie for all elements (c) is negatie for all elements (d) aries with the position of the element. Quick Quiz 18.4 Continuing with the scenario in Quick Quiz 18.3, at the moment the string has the shape shown by the red cure in Figure 18.9b, the instantaneous elocity of elements along the string (a) is zero for all elements (b) is positie for all elements (c) is negatie for all elements (d) aries with the position of the element. Example 18.2 Formation of a Standing Wae Two waes traeling in opposite directions produce a standing wae. The indiidual wae functions are y 1 (4.0 cm) sin(3.0x 2.0t) y 2 (4.0 cm) sin(3.0x 2.0t) where x and y are measured in centimeters. () Find the amplitude of the simple harmonic motion of the element of the medium located at x 2.3 cm. Solution The standing wae is described by Equation 18.3; in this problem, we hae 4.0 cm, k 3.0 rad/cm, and 2.0 rad/s. Thus, y (2 sin kx) cos t [(8.0 cm)sin 3.0x] cos 2.0t Thus, we obtain the amplitude of the simple harmonic motion of the element at the position x 2.3 cm by ealuating the coefficient of the cosine function at this position: y max (8.0 cm)sin 3.0x x2.3 (B) Find the positions of the nodes and antinodes if one end of the string is at x 0. Solution With k 2/ 3.0 rad/cm, we see that the waelength is (2/3.0) cm. Therefore, from Equation 18.4 we find that the nodes are located at x n (8.0 cm) sin (6.9 rad) 2 n 3 cm n 4.6 cm 0, 1, 2, 3,... and from Equation 18.5 we find that the antinodes are located at x n 4 n 6 cm n (C) What is the maximum alue of the position in the simple harmonic motion of an element located at an antinode? Solution ccording to Equation 18.3, the maximum position of an element at an antinode is the amplitude of the standing wae, which is twice the amplitude of the indiidual traeling waes: y max 2(sin kx) max 2(4.0 cm)( 1) where we hae used the fact that the maximum alue of sin kx is 1. Let us check this result by ealuating the coefficient of our standing-wae function at the positions we found for the antinodes: y max (8.0 cm)sin 3.0x xn(/6) (8.0 cm) sin 3.0n 6 rad 1, 3, 5, cm (8.0 cm) sin n 2 rad 8.0 cm In ealuating this expression, we hae used the fact that n is an odd integer; thus, the sine function is equal to 1, depending on the alue of n Standing Waes in a String Fixed at Both Ends Consider a string of length L fixed at both ends, as shown in Figure Standing waes are set up in the string by a continuous superposition of waes incident on and reflected from the ends. Note that there is a boundary condition for the waes on the

11 S ECTION 18.3 Standing Waes in a String Fixed at Both Ends 553 L f 2 (a) n = 2 (c) L = λ 2 N N f 1 f 3 n = 1 L = 1 λ 2 1 (b) n = 3 (d) L = 3 λ 2 3 ctie Figure (a) string of length L fixed at both ends. The normal modes of ibration form a harmonic series: (b) the fundamental, or first harmonic; (c) the second harmonic; (d) the third harmonic. t the ctie Figures link at you can choose the mode number and see the corresponding standing wae. string. The ends of the string, because they are fixed, must necessarily hae zero displacement and are, therefore, nodes by definition. The boundary condition results in the string haing a number of natural patterns of oscillation, called normal modes, each of which has a characteristic frequency that is easily calculated. This situation in which only certain frequencies of oscillation are allowed is called quantization. Quantization is a common occurrence when waes are subject to boundary conditions and will be a central feature in our discussions of quantum physics in the extended ersion of this text. Figure shows one of the normal modes of oscillation of a string fixed at both ends. Except for the nodes, which are always stationary, all elements of the string oscillate ertically with the same frequency but with different amplitudes of simple harmonic motion. Figure represents snapshots of the standing wae at arious times oer one half of a period. The red arrows show the elocities of arious elements of the string at arious times. s we found in Quick Quizzes 18.3 and 18.4, (a) N t = 0 N N (b) t = T/ 8 (c) t = T/4 (d) t = 3T/ 8 (e) t = T/ 2 Figure standing-wae pattern in a taut string. The fie snapshots were taken at interals of one eighth of the period. (a) t t 0, the string is momentarily at rest. (b) t t T/8, the string is in motion, as indicated by the red arrows, and different parts of the string moe in different directions with different speeds. (c) t t T/4, the string is moing but horizontal (undeformed). (d) The motion continues as indicated. (e) t t T/2, the string is again momentarily at rest, but the crests and troughs of (a) are reersed. The cycle continues until ultimately, when a time interal equal to T has passed, the configuration shown in (a) is repeated.

12 554 CHPTER 18 Superposition and Standing Waes all elements of the string hae zero elocity at the extreme positions (Figs a and 18.11e) and elements hae arying elocities at other positions (Figs b through 18.11d). The normal modes of oscillation for the string can be described by imposing the requirements that the ends be nodes and that the nodes and antinodes be separated by one fourth of a waelength. The first normal mode that is consistent with the boundary conditions, shown in Figure 18.10b, has nodes at its ends and one antinode in the middle. This is the longest-waelength mode that is consistent with our requirements. This first normal mode occurs when the length of the string is half the waelength 1, as indicated in Figure 18.10b, or 1 2L. The next normal mode (see Fig c) of waelength 2 occurs when the waelength equals the length of the string, that is, when 2 L. The third normal mode (see Fig d) corresponds to the case in which 3 2L/3. In general, the waelengths of the arious normal modes for a string of length L fixed at both ends are Waelengths of normal modes n 2L n n 1, 2, 3,... (18.6) Frequencies of normal modes as functions of wae speed and length of string Frequencies of normal modes as functions of string tension and linear mass density Fundamental frequency of a taut string where the index n refers to the nth normal mode of oscillation. These are the possible modes of oscillation for the string. The actual modes that are excited on a string are discussed shortly. The natural frequencies associated with these modes are obtained from the relationship f /, where the wae speed is the same for all frequencies. Using Equation 18.6, we find that the natural frequencies f n of the normal modes are f n n (18.7) These natural frequencies are also called the quantized frequencies associated with the ibrating string fixed at both ends. Because T/ (see Eq ), where T is the tension in the string and is its linear mass density, we can also express the natural frequencies of a taut string as f n n 2L n 2L T n 1, 2, 3, (18.8) The lowest frequency f 1, which corresponds to n 1, is called either the fundamental or the fundamental frequency and is gien by f 1 1 2L T (18.9) The frequencies of the remaining normal modes are integer multiples of the fundamental frequency. Frequencies of normal modes that exhibit an integer-multiple relationship such as this form a harmonic series, and the normal modes are called n 1, 2, 3, Multiflash photographs of standing-wae patterns in a cord drien by a ibrator at its left end. The single-loop pattern represents the first normal mode (the fundamental), n 1). The double-loop pattern represents the second normal mode (n 2), and the triple-loop pattern represents the third normal mode (n 3) Richard Megna/Fundamental Photographs

13 S ECTION 18.3 Standing Waes in a String Fixed at Both Ends 555 harmonics. The fundamental frequency f 1 is the frequency of the first harmonic; the frequency f 2 2f 1 is the frequency of the second harmonic; and the frequency f n nf 1 is the frequency of the nth harmonic. Other oscillating systems, such as a drumhead, exhibit normal modes, but the frequencies are not related as integer multiples of a fundamental. Thus, we do not use the term harmonic in association with these types of systems. In obtaining Equation 18.6, we used a technique based on the separation distance between nodes and antinodes. We can obtain this equation in an alternatie manner. Because we require that the string be fixed at x 0 and x L, the wae function y(x, t) gien by Equation 18.3 must be zero at these points for all times. That is, the boundary conditions require that y(0, t) 0 and y(l, t) 0 for all alues of t. Because the standing wae is described by y 2(sin kx) cos t, the first boundary condition, y(0, t) 0, is automatically satisfied because sin kx 0 at x 0. To meet the second boundary condition, y(l, t) 0, we require that sin kl 0. This condition is satisfied when the angle kl equals an integer multiple of rad. Therefore, the allowed alues of k are gien by 1 knl n n 1, 2, 3,... (18.10) Because k n 2/ n, we find that 2 n L n or n 2L n which is identical to Equation Let us examine further how these arious harmonics are created in a string. If we wish to excite just a single harmonic, we must distort the string in such a way that its distorted shape corresponds to that of the desired harmonic. fter being released, the string ibrates at the frequency of that harmonic. This maneuer is difficult to perform, howeer, and it is not how we excite a string of a musical instrument. If the string is distorted such that its distorted shape is not that of just one harmonic, the resulting ibration includes arious harmonics. Such a distortion occurs in musical instruments when the string is plucked (as in a guitar), bowed (as in a cello), or struck (as in a piano). When the string is distorted into a nonsinusoidal shape, only waes that satisfy the boundary conditions can persist on the string. These are the harmonics. The frequency of a string that defines the musical note that it plays is that of the fundamental. The frequency of the string can be aried by changing either the tension or the string s length. For example, the tension in guitar and iolin strings is aried by a screw adjustment mechanism or by tuning pegs located on the neck of the instrument. s the tension is increased, the frequency of the normal modes increases in accordance with Equation Once the instrument is tuned, players ary the frequency by moing their fingers along the neck, thereby changing the length of the oscillating portion of the string. s the length is shortened, the frequency increases because, as Equation 18.8 specifies, the normal-mode frequencies are inersely proportional to string length. Quick Quiz 18.5 When a standing wae is set up on a string fixed at both ends, (a) the number of nodes is equal to the number of antinodes (b) the waelength is equal to the length of the string diided by an integer (c) the frequency is equal to the number of nodes times the fundamental frequency (d) the shape of the string at any time is symmetric about the midpoint of the string. 1 We exclude n 0 because this alue corresponds to the triial case in which no wae exists (k 0).

14 556 CHPTER 18 Superposition and Standing Waes Example 18.3 Gie Me a C Note! Middle C on a piano has a fundamental frequency of 262 Hz, and the first aboe middle C has a fundamental frequency of 440 Hz. () Calculate the frequencies of the next two harmonics of the C string. Solution Knowing that the frequencies of higher harmonics are integer multiples of the fundamental frequency f Hz, we find that f 2 2f 1 f 3 3f 1 (B) If the and C strings hae the same linear mass density and length L, determine the ratio of tensions in the two strings. Solution Using Equation 18.9 for the two strings ibrating at their fundamental frequencies gies f 1 1 2L T 524 Hz 786 Hz and f 1C 1 2L T C Setting up the ratio of these frequencies, we find that f 1 f 1C T T C T T C f 1 f 1C What If? What if we look inside a real piano? In this case, the assumption we made in part (B) is only partially true. The string densities are equal, but the length of the string is only 64 percent of the length of the C string. What is the ratio of their tensions? nswer Using Equation 18.8 again, we set up the ratio of frequencies: f 1 f 1C L C L T T C T T C T (0.64) 2 T C Example 18.4 Guitar Basics Interactie The high E string on a guitar measures 64.0 cm in length and has a fundamental frequency of 330 Hz. By pressing down so that the string is in contact with the first fret (Fig ), the string is shortened so that it plays an F note that has a frequency of 350 Hz. How far is the fret from the neck end of the string? Solution Equation 18.7 relates the string s length to the fundamental frequency. With n 1, we can sole for the speed of the wae on the string, 2L n f n 2(0.640 m) 1 (330 Hz) 422 m/s Because we hae not adjusted the tuning peg, the tension in the string, and hence the wae speed, remain constant. We can again use Equation 18.7, this time soling for L and Charles D. Winters Figure (Example 18.4) Playing an F note on a guitar. substituting the new frequency to find the shortened string length: L n 2f n (1) The difference between this length and the measured length of 64.0 cm is the distance from the fret to the neck end of the string, or 3.7 cm. What If? What if we wish to play an F sharp, which we do by pressing down on the second fret from the neck in Figure 18.12? The frequency of F sharp is 370 Hz. Is this fret another 3.7 cm from the neck? nswer If you inspect a guitar fingerboard, you will find that the frets are not equally spaced. They are far apart near the neck and close together near the opposite end. Consequently, from this obseration, we would not expect the F sharp fret to be another 3.7 cm from the end. Let us repeat the calculation of the string length, this time for the frequency of F sharp: L n 2f n (1) 422 m/s 2(350 Hz) 422 m/s 2(370 Hz) m 60.3 cm m This gies a distance of m m m 6.9 cm from the neck. Subtracting the distance from the neck to the first fret, the separation distance between the first and second frets is 6.9 cm 3.7 cm 3.2 cm. Explore this situation at the Interactie Worked Example link at

15 S ECTION 18.3 Standing Waes in a String Fixed at Both Ends 557 Example 18.5 Changing String Vibration with Water Interactie One end of a horizontal string is attached to a ibrating blade and the other end passes oer a pulley as in Figure 18.13a. sphere of mass 2.00 kg hangs on the end of the string. The string is ibrating in its second harmonic. container of water is raised under the sphere so that the sphere is completely submerged. fter this is done, the string ibrates in its fifth harmonic, as shown in Figure 18.13b. What is the radius of the sphere? Solution To conceptualize the problem, imagine what happens when the sphere is immersed in the water. The buoyant force acts upward on the sphere, reducing the tension in the string. The change in tension causes a change in the speed of waes on the string, which in turn causes a change in the waelength. This altered waelength results in the string ibrating in its fifth normal mode rather than the second. We categorize the problem as one in which we will need to combine our understanding of Newton s second law, buoyant forces, and standing waes on strings. We begin to analyze the problem by studying Figure 18.13a. Newton s second law applied to the sphere tells us that the tension in the string is equal to the weight of the sphere: F T 1 mg 0 T 1 mg (2.00 kg)(9.80 m/s 2 ) 19.6 N where the subscript 1 is used to indicate initial ariables before we immerse the sphere in water. Once the sphere is immersed in water, the tension in the string decreases to T 2. pplying Newton s second law to the sphere again in this situation, we hae T 2 B mg 0 (1) B mg T 2 The desired quantity, the radius of the sphere, will appear in the expression for the buoyant force B. Before proceeding in this direction, howeer, we must ealuate T 2. We do this from the standing wae information. We write the equation for the frequency of a standing wae on a string (Equation 18.8) twice, once before we immerse the sphere and once after, and diide the equations: f n 1 2L T 1 f n 2 2L T 2 : 1 n 1 n 2 T 1 T 2 where the frequency f is the same in both cases, because it is determined by the ibrating blade. In addition, the linear mass density and the length L of the ibrating portion of the string are the same in both cases. Soling for T 2, we hae T 2 n 1 n 2 2 T (19.6 N) 3.14 N Substituting this into Equation (1), we can ealuate the buoyant force on the sphere: B mg T N 3.14 N 16.5 N Finally, expressing the buoyant force (Eq. 14.5) in terms of the radius of the sphere, we sole for the radius: 4 3 B water gv sphere water g ( r 3 ) (a) r 3 3B 4waterg 3 3(16.5 N) 4(1 000 kg/m 3 )(9.80 m/s 2 ) m 7.38 cm Figure (Example 18.5) When the sphere hangs in air, the string ibrates in its second harmonic. When the sphere is immersed in water, the string ibrates in its fifth harmonic. (b) To finalize this problem, note that only certain radii of the sphere will result in the string ibrating in a normal mode. This is because the speed of waes on the string must be changed to a alue such that the length of the string is an integer multiple of half waelengths. This is a feature of the quantization that we introduced earlier in this chapter the sphere radii that cause the string to ibrate in a normal mode are quantized. You can adjust the mass at the Interactie Worked Example link at

16 558 CHPTER 18 Superposition and Standing Waes 18.4 Resonance mplitude f 0 Frequency of driing force Figure Graph of the amplitude (response) ersus driing frequency for an oscillating system. The amplitude is a maximum at the resonance frequency f 0. We hae seen that a system such as a taut string is capable of oscillating in one or more normal modes of oscillation. If a periodic force is applied to such a system, the amplitude of the resulting motion is greatest when the frequency of the applied force is equal to one of the natural frequencies of the system. We discussed this phenomenon, known as resonance, briefly in Section lthough a block spring system or a simple pendulum has only one natural frequency, standingwae systems hae a whole set of natural frequencies, such as that gien by Equation 18.7 for a string. Because an oscillating system exhibits a large amplitude when drien at any of its natural frequencies, these frequencies are often referred to as resonance frequencies. Figure shows the response of an oscillating system to arious driing frequencies, where one of the resonance frequencies of the system is denoted by f 0. Note that the amplitude of oscillation of the system is greatest when the frequency of the driing force equals the resonance frequency. The maximum amplitude is limited by friction in the system. If a driing force does work on an oscillating system that is initially at rest, the input energy is used both to increase the amplitude of the oscillation and to oercome the friction force. Once maximum amplitude is reached, the work done by the driing force is used only to compensate for mechanical energy loss due to friction. B Figure n example of resonance. If pendulum is set into oscillation, only pendulum C, whose length matches that of, eentually oscillates with large amplitude, or resonates. The arrows indicate motion in a plane perpendicular to the page. Vibrating blade Figure Standing waes are set up in a string when one end is connected to a ibrating blade. When the blade ibrates at one of the natural frequencies of the string, large-amplitude standing waes are created. C D Examples of Resonance playground swing is a pendulum haing a natural frequency that depends on its length. Wheneer we use a series of regular impulses to push a child in a swing, the swing goes higher if the frequency of the periodic force equals the natural frequency of the swing. We can demonstrate a similar effect by suspending pendulums of different lengths from a horizontal support, as shown in Figure If pendulum is set into oscillation, the other pendulums begin to oscillate as a result of waes transmitted along the beam. Howeer, pendulum C, the length of which is close to the length of, oscillates with a much greater amplitude than pendulums B and D, the lengths of which are much different from that of pendulum. Pendulum C moes the way it does because its natural frequency is nearly the same as the driing frequency associated with pendulum. Next, consider a taut string fixed at one end and connected at the opposite end to an oscillating blade, as illustrated in Figure The fixed end is a node, and the end connected to the blade is ery nearly a node because the amplitude of the blade s motion is small compared with that of the elements of the string. s the blade oscillates, transerse waes sent down the string are reflected from the fixed end. s we learned in Section 18.3, the string has natural frequencies that are determined by its length, tension, and linear mass density (see Eq. 18.8). When the frequency of the blade equals one of the natural frequencies of the string, standing waes are produced and the string oscillates with a large amplitude. In this resonance case, the wae generated by the oscillating blade is in phase with the reflected wae, and the string absorbs energy from the blade. If the string is drien at a frequency that is not one of its natural frequencies, then the oscillations are of low amplitude and exhibit no stable pattern. Once the amplitude of the standing-wae oscillations is a maximum, the mechanical energy deliered by the blade and absorbed by the system is transformed to internal energy because of the damping forces caused by friction in the system. If the applied frequency differs from one of the natural frequencies, energy is transferred to the string at first, but later the phase of the wae becomes such that it forces the blade to receie energy from the string, thereby reducing the energy in the string. Resonance is ery important in the excitation of musical instruments based on air columns. We shall discuss this application of resonance in Section 18.5.

17 SECTION Standing Waes in ir Columns 559 Quick Quiz 18.6 wine glass can be shattered through resonance by maintaining a certain frequency of a high-intensity sound wae. Figure 18.17a shows a side iew of a wine glass ibrating in response to such a sound wae. Sketch the standingwae pattern in the rim of the glass as seen from aboe. If an integral number of waes fit around the circumference of the ibrating rim, how many waelengths fit around the rim in Figure 18.17a? Courtesy of Professor Thomas D. Rossing, Northern Illinois Uniersity 1992 Ben Rose/The Image Bank (a) Figure (Quick Quiz 18.6) (a) Standing-wae pattern in a ibrating wine glass. The glass shatters if the amplitude of ibration becomes too great. (b) wine glass shattered by the amplified sound of a human oice. (b) 18.5 Standing Waes in ir Columns Standing waes can be set up in a tube of air, such as that inside an organ pipe, as the result of interference between longitudinal sound waes traeling in opposite directions. The phase relationship between the incident wae and the wae reflected from one end of the pipe depends on whether that end is open or closed. This relationship is analogous to the phase relationships between incident and reflected transerse waes at the end of a string when the end is either fixed or free to moe (see Figs and 16.15). In a pipe closed at one end, the closed end is a displacement node because the wall at this end does not allow longitudinal motion of the air. s a result, at a closed end of a pipe, the reflected sound wae is 180 out of phase with the incident wae. Furthermore, because the pressure wae is 90 out of phase with the displacement wae (see Section 17.2), the closed end of an air column corresponds to a pressure antinode (that is, a point of maximum pressure ariation). The open end of an air column is approximately a displacement antinode 2 and a pressure node. We can understand why no pressure ariation occurs at an open end by noting that the end of the air column is open to the atmosphere; thus, the pressure at this end must remain constant at atmospheric pressure. You may wonder how a sound wae can reflect from an open end, as there may not appear to be a change in the medium at this point. It is indeed true that the medium 2 Strictly speaking, the open end of an air column is not exactly a displacement antinode. compression reaching an open end does not reflect until it passes beyond the end. For a tube of circular cross section, an end correction equal to approximately 0.6R, where R is the tube s radius, must be added to the length of the air column. Hence, the effectie length of the air column is longer than the true length L. We ignore this end correction in this discussion.

18 560 CHPTER 18 Superposition and Standing Waes through which the sound wae moes is air both inside and outside the pipe. Howeer, sound is a pressure wae, and a compression region of the sound wae is constrained by the sides of the pipe as long as the region is inside the pipe. s the compression region exits at the open end of the pipe, the constraint of the pipe is remoed and the compressed air is free to expand into the atmosphere. Thus, there is a change in the character of the medium between the inside of the pipe and the outside een though there is no change in the material of the medium. This change in character is sufficient to allow some reflection. With the boundary conditions of nodes or antinodes at the ends of the air column, we hae a set of normal modes of oscillation, as we do for the string fixed at both ends. Thus, the air column has quantized frequencies. The first three normal modes of oscillation of a pipe open at both ends are shown in Figure 18.18a. Note that both ends are displacement antinodes (approximately). In the first normal mode, the standing wae extends between two adjacent antinodes, which is a distance of half a waelength. Thus, the waelength is twice the length of the pipe, and the fundamental frequency is f 1 /2L. s Figure 18.18a shows, the frequencies of the higher harmonics are 2f 1, 3f 1,.... Thus, we can say that In a pipe open at both ends, the natural frequencies of oscillation form a harmonic series that includes all integral multiples of the fundamental frequency. PITFLL PREVENTION 18.3 Sound Waes in ir re Longitudinal, not Transerse Note that the standing longitudinal waes are drawn as transerse waes in Figure This is because it is difficult to draw longitudinal displacements they are in the same direction as the propagation. Thus, it is best to interpret the cures in Figure as a graphical representation of the waes (our diagrams of string waes are pictorial representations), with the ertical axis representing horizontal displacement of the elements of the medium. N N N N N N N N N N N N L (a) Open at both ends λ 1 = 2L f 1 = = λ 1 2L λ 2 = L f 2 = = 2f L 1 2 λ 3 = L 3 f 3 = 3 = 3f 2L 1 λ 1 = 4L f 1 = = λ 1 4L 4 λ 3 = L 3 f 3 = 3 = 3f 4L 1 4 λ 5 = L 5 f 5 = 5 = 5f 4L 1 First harmonic Second harmonic Third harmonic First harmonic Third harmonic Fifth harmonic (b) Closed at one end, open at the other Figure Motion of elements of air in standing longitudinal waes in a pipe, along with schematic representations of the waes. In the schematic representations, the structure at the left end has the purpose of exciting the air column into a normal mode. The hole in the upper edge of the column assures that the left end acts as an open end. The graphs represent the displacement amplitudes, not the pressure amplitudes. (a) In a pipe open at both ends, the harmonic series created consists of all integer multiples of the fundamental frequency: f 1, 2f 1, 3f 1,.... (b) In a pipe closed at one end and open at the other, the harmonic series created consists of only odd-integer multiples of the fundamental frequency: f 1, 3f 1, 5f 1,....