Solution to HW15 CJ5 17.CQ.003. REASONING AND SOLUTION Consider the situation in Figure 17.3. If you walk along a line that is perpendicular to the line between the speakers and passes through the overlap point, you will always be equidistant from both speakers. Therefore the sound waves along this line always overlap exactly in phase. Hence, you will always hear the same loudness; you will not observe the loudness to change from loud to faint to loud. On the other hand, if you walk along a line that passes through the overlap point and is parallel to the line between the speakers, your distance from the two speakers will vary such that the difference in path lengths traveled by the two waves will vary. At certain points the path length difference between the two sound waves will be an integer number of wavelengths; constructive interference will occur at these points, and the sound intensity will be a maximum. At other points, the path length difference between the two sound waves will be an odd number of half-wavelengths [(1/2)λ, (3/2)λ, (5/2)λ, etc.]; destructive interference will occur and the sound intensity will be a minimum. In between the points of constructive and destructive interference, the waves will be out of phase by varying degrees. Therefore, as you walk along this line, you will observe the sound intensity to alternate between faint and loud. CJ5 17.CQ.005. REASONING AND SOLUTION When a wave encounters an obstacle or the edges of an opening, it bends or diffracts around them. The extent of the diffraction depends on the ratio of the wavelength λ to the size D of the obstacle or opening. If the ratio λ / D is small, little diffraction occurs. As the ratio λ / D is made larger, the wave diffracts to a greater extent. In Example 1 in Section 16.2, it is shown that the wavelength of AM radio waves is 244 m, while the wavelength of FM radio waves is 3.26 m. For a given obstacle, the ratio λ / D will be greater for AM radio waves; therefore AM radio waves will diffract more readily around a given obstacle than FM waves. CJ5 17.CQ.006. REASONING AND SOLUTION A tuning fork has a frequency of 440 Hz. The string of a violin and this tuning fork, when sounded together, produce a beat frequency of 1 Hz. This beat frequency is the difference between the frequency of the tuning fork and the frequency of the violin string. A violin string of frequency 439 Hz, as well as a violin string of 441 Hz, will produce a beat frequency of 1 Hz, when sounded together with the 440 Hz tuning fork. We conclude that, from these two pieces of information alone, it is not possible to distinguish between these two possibilities. Therefore, it is not possible to determine the frequency of the violin string. CJ5 17.CQ.009. REASONING AND SOLUTION A string is attached to a wall and vibrates back and forth as in Figure 17.18. The vibration frequency and length of the string are fixed. The tension in the string is changed. From Equation 16.2, vF= /(m/L), we see that increasing the tension F results in increasing the speed v of the waves on the string. From the relationship v = λ f , we see that, since the frequency remains fixed, an increase in the wave speed results in an increase in the wavelength of the wave. It is observed that at certain values of the tension, a standing wave pattern develops. Since the two ends of the string are "fixed," the ends of the string are nodes; thus, the length L of the string must contain an integer number of half- λ = wavelengths. Therefore, standing waves will occur at the wavelengths n 2L/ n, where n = 1, 2, 3, 4, . . . The largest possible wavelength that will result in a standing wave pattern occurs when n = 1, and the wavelength is equal to twice the length of the string. If the tension is increased beyond the value for which λ = 2L, the string cannot sustain a standing wave pattern.

CJ5 17.CQ.013. REASONING AND SOLUTION Consider the tube at the bottom of Figure 17.23. We suppose that the air in the tube is replaced with a gas in which the speed of sound is twice what it is in air. If the frequency of the tuning fork remains unchanged, we see from the relation v = λ f that the wavelength of the sound will be twice as long in the gas as it is in air. The resulting pattern is shown in the drawing at the right.

CJ5 17.P.007. WWW WWW REASONING The geometry of the positions of the loudspeakers and the listener is shown in the following drawing.

C

d = 1.00 m 1 d2 y

60.0° A B x x 1 2

The listener at C will hear either a loud sound or no sound, depending upon whether the interference occurring at C is constructive or destructive. If the listener hears no sound, destructive interference occurs, so nλ dd−= n =135,,, 212 K (1)

SOLUTION Sincev= λ f, according to Equation 16.1, the wavelength of the tone is

v 343 m / s λ = = = 5.00 m f 68.6 Hz

Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so

nλ 5.00 m d =+d = +1.00 m = 3.50 m 212 2

From the figure above we have that,

=°= x1 (1.00 m) cos 60.0 0.500 m

y =°(1.00 m) sin 60.0 =0.866 m Then

22+=2 = 2 =−22= xy2 d2 (.350 m) or x2 (.3 50 m) (0.866 m) 3.39 m

Therefore, the closest that speaker A can be to speaker B so that the listener hears no += + = sound is xx120.500 m 3.39 m 3.89 m . CJ5 17.P.011. SSM REASONING The diffraction angle for the first minimum for a circular opening is given by Equation 17.2: sinθ =1.22λ / D, where D is the diameter of the opening.

SOLUTION a. Using Equation 16.1, we must first find the wavelength of the 2.0-kHz tone:

v 343 m/s λ = = = 0.17 m f 2.0 ×103 Hz

The diffraction angle for a 2.0-kHz tone is, therefore,

 0.17 m  θ = sin –1  1.22 ×  = 44°  0.30 m 

b. The wavelength of a 6.0-kHz tone is

v 343 m/s λ = = = 0.057 m f 6.0 ×103 Hz

Therefore, if we wish to generate a 6.0-kHz tone whose diffraction angle is as wide as that for the 2.0-kHz tone in part (a), we will need a speaker of diameter D, where

1.22 λ (1.22)(0.057 m) D = = = 0.10 m sinθ sin 44° CJ5 17.P.015. REASONING AND Diffraction horn SOLUTION Stage The figure at the right shows the geometry of the situation.

8.7 m θ θ The tone will not be heard at seats located at the first diffraction minimum. This occurs when First row of seats C λ v sin θ = = x D fD That is, the angle θ is given by

−−F v I L 343 m / s O θ =sin 11= sin =°27.2 G J M 4 P H fDK N (1.0 ×10 Hz)(0.075 m) Q

From the figure at the right, we see that

x tan 27.2°= ⇒ x = (8.7 m)(tan 27.2°) = 4.47 m 8.7 m θ 8.7 m

x Thus, seats at which the tone cannot be heard are a distance x on either side of the center seat C. Thus, the distance between the two seats is

2x = 2(4.47 m) = 8.9 m

CJ5 17.P.017. SSM REASONING AND SOLUTION Two ultrasonic sound waves combine and form a beat frequency that is in the range of human hearing. The frequency of one of the ultrasonic waves is 70 kHz. The beat frequency is the difference between the two sound frequencies. The smallest possible value for the ultrasonic frequency can be found by subtracting the upper limit of human hearing from the value of 70 kHz. The largest possible value for the ultrasonic frequency can be determined by adding the upper limit of human hearing to the value of 70 kHz. We know that the frequency range of human hearing is from 20 Hz to 20 kHz.

a. The smallest possible frequency of the other ultrasonic wave is

f = 70 kHz – 20 kHz = 50 kHz

which results in a beat frequency of 70 kHz – 50 kHz = 20 kHz .

b. The largest possible frequency for the other wave is

f = 70 kHz + 20 kHz = 90 kHz

which results in a beat frequency of 90 kHz – 70 kHz = 20 kHz . CJ5 17.P.025. REASONING AND SOLUTION Assuming that fretting the string does NOT change the tension, the speed of waves on the string will be the same in both cases. The speed of the waves is given v = λf = 2Lf. Applied to the first case (unfretted), this relation gives v = 2L1f1. Applied to the second case, it gives v = 2L2f2.

Equating the above equations and rearranging yields

L2 = L1(f1/f2) = (0.62 m)(196 Hz)/(262 Hz) = 0.46 m