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CHAPTER

15

Wave Motion

1* · A rope hangs vertically from the ceiling. Do waves on the rope move faster, slower, or at the same speed as they move from bottom to top? Explain. They move faster as they move up because the tension increases due to the weight of the rope below. 2

· (a) The bulk modulus for water is 2.0 × 109 N/m2. Use it to find the speed of sound in water. (b)The speed of sound in mercury is 1410 m/s. What is the bulk modulus for mercury (ρ = 13.6 ×103 kg/m3)? (a) Use Equ. 15-4 v = 2 ×10 9 / 103 m/s = 1.41× 103 m/s 2 (b) From Equ. 15-4, B = v ρ B = (14102 × 1.36 × 103) N/m2 = 2.70 × 1010 N/m2

3

· Calculate the speed of sound waves in hydrogen gas at T = 300 K. (Take M = 2 g/mol and γ = 1.4.) Use Equ. 15-5; M = 2 × 10−3 kg/mol v = 1.4 × 8.314 × 300/2 ×10 −3 m/s = 1320 m/s

4

· A steel wire 7 m long has a mass of 100 g. It is under a tension of 900 N. What is the speed of a transverse wave pulse on this wire? Use Equ. 15-3 900 v= m/s = 251 m/s

0.1/7.0

5* · Transverse waves travel at 150 m/s on a wire of length 80 cm that is under a tension of 550 N. What is the mass of the wire? From Equ. 15-3, µ = F/v2; m = µL = FL/v2 m = (550 × 0.8/1502) kg = 0.0196 kg = 19.6 g 6

· A wave pulse propagates along a wire in the positive x direction at 20 m/s. What will the pulse velocity be if we (a) double the length of the wire but keep the tension and mass per unit length constant? (b) double the tension while holding the length and mass per unit length constant? (c) double the mass per unit length while holding the other variables constant? See Equ. 15-3. (a) 20 m/s. (b) 20 2 m/s = 28.8 m/s. (c) 20/ 2 m/s = 14.1 m/s.

Chapter 15 7

Wave Motion

· A steel piano wire is 0.7 m long and has a mass of 5 g. It is stretched with a tension of 500 N. (a) What is the speed of transverse waves on the wire? (b) To reduce the wave speed by a factor of 2 without changing the tension, what mass of copper wire would have to be wrapped around the steel wire? (a) Use Equ. 15-3 500 v= m/s = 265 m/s

0.005/0.7

(b) From Equ. 15-3, mf = 4mi

8

∆m = 3mi = 15 g

· The cable of a ski lift runs 400 m up a mountain and has a mass of 80 kg. When the cable is struck with a transverse blow at one end, the return pulse is detected 12 s later. (a) What is the speed of the wave? (b) What is the tension in the cable? v = ∆x/∆t v = 800/12 m/s = 66.7 m/s 2 From Equ. 15-3, F = v m/L F = (66.72 × 80/400) N = 889 N

9* ·· A common method for estimating the distance to a lightning flash is to begin counting when the flash is observed and continue until the thunder clap is heard. The number of seconds counted is then divided by 3 to get the distance in kilometers. (a) What is the velocity of sound in kilometers per second? (b) How accurate is this procedure? (c) Is a correction for the time it takes for the light to reach you important? (The speed of light is 3 × 108 m/s.) (a) v = 340 m/s = 0.340 km/s. (b) s = vt = 0.340t ≈ t/3 = 0.333t; error = 7/340 = 2%. (c) No; e.g., if t = 3 s, s ≈ 1 km, and the time required for light to travel 1 km is only 10−5 s. 10 ·· A method for measuring the speed of sound using an ordinary watch with a second hand is to stand some distance from a large flat wall and clap your hands rhythmically in such a way that the echo from the wall is heard halfway between every two claps. (a) Show that the speed of sound is given by v = 4LN, where L is the distance to the wall and N is the number of claps per second. (b) What is a reasonable value for L for this experiment to be feasible? (If you have access to a flat wall outdoors somewhere, try this method and compare your result with the standard value for the speed of sound.) (a) Time of travel = ∆t = 2L/v = 1/2(1/N). So v = 4LN. (b) Assume N = 1 clap/s; then L = 340/4 m = 85 m. So a distance of about 80 m is appropriate. 11 ·· A man drops a stone from a high bridge and hears it strike the water below exactly 4 s later. (a) Estimate the distance to the water based on the assumption that the travel time for the sound to reach the man is negligible. (b) Improve your estimate by using your result from part (a) for the distance to the water to estimate the time it takes for sound to travel this distance and then calculate the distance the rock falls in 4 s minus this time. (c) Calculate the exact distance and compare your result with your previous estimates. (a) d = 1/2gt2 d = 1/2 × 9.81 × 16 m = 78.5 m (b) 1. ∆t = d/vs ∆t = 78.5/340 s = 0.23 s 2 2. d ′ = 1/2g(t − ∆t) d ′ = 1/2 × 9.81 × 3.772 m = 69.7 m For t = 4 s, vs = 340 m/s, d = 70.5 m (c) t = 2 d/g + d/vs; 2d/g = t2 − 2dt/vs + d 2/vs2

Chapter 15 12 ··

Wave Motion

(a) Compute the derivative of the speed of a wave on a string with respect to the tension dv/dF, and show that the 1

differentials dv and dF obey dv/v = 2 dF/F. (b) A wave moves with a speed of 300 m/s on a wire that is under a tension of 500 N. Using dF to approximate a change in tension, determine how much the tension must be changed to increase the speed to 312 m/s.

(a)

dv d = dF dF

F 1 = µ 2

1 1 v dv 1 dF = ; = . Fµ 2 F v 2 F

(b) ∆F = 2F(∆v/v) = 2 × 500 × 12/300 N = 40 N. 13* ·· (a) Compute the derivative of the velocity of sound with respect to the absolute temperature, and show that the 1 differentials dv and dT obey dv/v = 2 dT/T . (b) Use this result to compute the percentage change in the velocity of

sound when the temperature changes from 0 to 27°C. (c) If the speed of sound is 331 m/s at 0°C, what is it (approximately) at 27°C? How does this approximation compare with the result of an exact calculation? (a) Follow the same procedure as in Problem 12. Since v ∝ T , dv/v = 1/2dT/T. (b) ∆T/T = 27/273; ∆v/v = 4.95%. (c) v300 = v273(1.0495) = 347 m/s. Using the fact that v ∝ T we obtain v 300 = 331 300/273 m/s = 347 m/s. 14 ··· In this problem, you will derive a convenient formula for the speed of sound in air at temperature t in Celsius degrees. Begin by writing the temperature as T = T0 + ∆T, where T0 = 273 K corresponds to 0°C and ∆T = t, the Celsius temperature. The speed of sound is a function of T, v(T). To a first-order approximation, you can write

v(T) ≈ v( T 0 ) + (dv/dT )T 0 ∆T , where (dv/dT )T 0 is the derivative evaluated at T = T0. Compute this derivative, and show that the result leads to

 t  v = ( 331 m/s )1 +  = (331 + 0.606t ) m/s  2T 0  Since t differs from T only by an additive constant, dv/dt = 1/2(v/T) = 1/2[v/(t + 273)]. So v(t) = v(0o C) + ∆v, where ∆v = 1/2(331 m/s)[t/(t + 273)]. For t > L0. Then v = L k/m . 3 Show explicitly that the following functions satisfy the wave equation: (a) y(x,t) = k(x+ vt ) ; 0(b)

17* ·

y(x,t) = Aeik(x- vt) , where A and k are constants and i = - 1 ; and (c) y(x,t) = ln k(x − vt).

(a) ∂ y/∂ x = 3k(x + vt)2; ∂ 2y/∂ x2 = 6k(x + vt); ∂ y/∂ t = 3kv(x + vt)2; ∂ 2y/∂ t2 = 6kv2(x + vt). v2(∂ 2y/∂ x2) = ∂ 2y/∂ t2. (b) ∂ y/∂ x = ikAe ik(x - vt); ∂ 2y/∂ x2 = −k 2Ae ik(x - vt); ∂ y/∂ t = −ikvAe ik(x - vt); ∂ 2y/∂ t2 = −k 2v2Ae ik(x - vt) = v2(∂ 2y/∂ x2). (c) ∂ y/∂ x = 1/(x − vt); ∂ 2y/∂ x2 = −1/(x − vt)2; ∂ y/∂ t = −v/(x − vt); ∂ 2y/∂ t2 = −v2/(x - vt)2 = v2(∂ 2y/∂ x2). Show that the function y = A sin kx cos ωt satisfies the wave equation. 2 2 2 2 2 2 2 2 2 2 ∂ y/∂ x = −Ak sin kx cos ωt; ∂ y/∂ t = −Aω sin kx cos ωt. ∂ y/∂ x = (1/v )∂ y/∂ t if v = ω/k. 19 ··· Consider the following equation: 18 ·

2

2 ∂ y + i α ∂y = 0 , i = ∂ x2 ∂t

−1

where α is a constant. Show that y(x, t) = A sin(kx − ωt) is not a solution of this equation but that the functions y(x,t) = Aei(kx - ω t) and y(x,t) = Aei(kx+ ωt) do satisfy that equation.

1. For y(x, t) = A sin(kx − ωt), ∂ 2y/∂ x2 = −Ak 2 sin(kx − ωt) and i(∂ y/∂ t) = −Aiω cos(kx − ωt). Evidently, 2 2 ∂ y/∂ x + i(∂ y/∂ t) ≠ 0. 2. For y(x, t) = Ae i(kx - ωt), ∂ 2y/∂ x2 = −k 2y and i(∂ y/∂ t) = ωy. Likewise, for y(x,t) = Ae i(kx + ωt). Both functions are solutions of the equation provided k 2 = αω. 20 · A traveling wave passes a point of observation. At this point, the time between successive crests is 0.2 s. Which of the following is true?

Chapter 15

Wave Motion

(a) The wavelength is 5 m. (b) The frequency is 5 Hz. (c) The velocity of propagation is 5 m/s. (d) The wavelength is 0.2 m. (e) There is not enough information to justify any of these statements. (b) is correct; T = 1/f. 21* · True or false: The energy in a wave is proportional to the square of the amplitude of the wave. True; see Equ. 15-24. 22 · A rope hangs vertically. You shake the bottom back and forth, creating a sinusoidal wave train. Is the wavelength at the top the same as, less than, or greater than the wavelength at the bottom? The wavelength is greater at the top; λ = v/f, f is constant and v is greater (see Problem 1). 23 · One end of a string 6 m long is moved up and down with simple harmonic motion at a frequency of 60 Hz. The waves reach the other end of the string in 0.5 s. Find the wavelength of the waves on the string. 1. Find the wave velocity 2. λ = v/f

v = (6/0.5) m/s = 12 m/s λ = 12/60 m = 20 cm

24 · Equation 15-13 expresses the displacement of a harmonic wave as a function of x and t in terms of the wave parameters k and ω. Write the equivalent expressions that contain the following pairs of parameters instead of k and ω: (a) k and v, (b) λ and f, (c) λ and T, (d) λ and v, and (e) f and v. We have the following relations: k = 2π /λ; ω = 2π f; v = f λ so ω = 2π v/λ = kv. For a wave traveling to the right, the expressions are then: (a) y = A sin k(x − vt); (b) y = A sin 2π (x/λ − ft); (c) y = A sin 2π (x/λ − t/T); (d) y = A sin[(2π /λ)(x − vt)]; (e) y = A sin 2π f(x/v − t). For a wave traveling to the left, change the − to a + sign. 25* · Equation 15-10 applies to all types of periodic waves, including electromagnetic waves such as light waves and microwaves, which travel at 3 × 108 m/s in a vacuum. (a) The range of wavelengths of light to which the eye is sensitive is about 4 × 10– 7 to 7 × 10– 7 m. What are the frequencies that correspond to these wavelengths? (b) Find the frequency of a microwave that has a wavelength of 3 cm. (a) f = v/λ; v = c = 3 × 108 m/s f min =(3 × 108/7 × 10−7) Hz ≈ 4.3 × 1014 Hz; f max = 7.5 × 1014 Hz (b) f = c/λ f = 3 × 108/3 × 10−2 Hz = 1010 Hz 26 · A harmonic wave on a string with a mass per unit length of 0.05 kg/m and a tension of 80 N has an amplitude of 5 cm. Each section of the string moves with simple harmonic motion at a frequency of 10 Hz. Find the power propagated along the string. Find v = (F/µ)1/2 v = 80/0.05 m/s = 40 m/s P = 1/2µω2A2v; ω = 2π f P = (0.05 × 400π 2 × 25 × 10−4 × 40)/2 W = 9.87 W 27 ·

A rope 2 m long has a mass of 0.1 kg. The tension is 60 N. A power source at one end sends a harmonic wave

304

Chapter 15

Wave Motion

with an amplitude of 1 cm down the rope. The wave is extracted at the other end without any reflection. What is the frequency of the power source if the power transmitted is 100 W? 1. Determine the wave velocity v = F/µ = 60/(0.1/2) m/s = 34.6 m/s 2. From Equation 15-19, f =

1 2πA

2P µv

f = [(1/2π × 0.01)(200/34.6 × 0.05) 1/2] Hz = 171 Hz

28 ·· The wave function for a harmonic wave on a string is y(x, t) = (0.001 m) sin(62.8 m−1x + 314s−1t). (a) In what direction does this wave travel, and what is its speed? (b) Find the wavelength, frequency, and period of this wave. (c) What is the maximum speed of any string segment? (a) The wave travels to the left (−x direction); v = ω/k = 314/62.8 m/s = 5.0 m/s. (b) λ = 2π /k = 0.1 m; f = ω/2π = 50 Hz; T = 1/f = 0.02 s. (c) vmax = Aω = 0.314 m/s. 29* ·· A harmonic wave with a frequency of 80 Hz and an amplitude of 0.025 m travels along a string to the right with a speed of 12 m/s. (a) Write a suitable wave function for this wave. (b) Find the maximum speed of a point on the string. (c) Find the maximum acceleration of a point on the string. (a) See Problem 24(e). y(x, t) = 0.025 sin[(160π )(x/12 − t)] m = 0.025 sin(41.9x − 503t) m. (b) vmax = Aω = (0.025 × 503) m/s = 12.6 m/s. (c) a max = Aω 2 = ωvmax = 6321 m/s2. 30 ·· Waves of frequency 200 Hz and amplitude 1.2 cm move along a 20-m string that has a mass of 0.06 kg and a tension of 50 N. (a) What is the average total energy of the waves on the string? (b) Find the power transmitted past a given point on the string. (a) Apply Equ. 15-18 with ∆x = L E = 1/2[(0.06/20)(4π 2 × 2002)(0.0122)(20)] J = 6.82 J (b) 1. Determine v = F/ µ 2. Evaluate P using Equs. 15-19 and 15-18

v = 50/(0.06/2 0) m/s = 129 m/s P = Ev/L = (6.82 × 129/20) W = 44 W

31 ·· In a real string, a wave loses some energy as it travels down the string. Such a situation can be described by a wave function whose amplitude A(x) depends on x:

y = A(x)sin (kx − ωt) = ( A0 e − bx ) sin (kx − ωt) (a) What is the original power carried by the wave at the origin? (b) What is the power transported by the wave at point x? (a) At x = 0, A = A0; P(0) = 1/2µω2A02v. (b) At x, A2 = A02e−2bx; P(x) = 1/2µω2A02ve−2bx. 32 ·· Power is to be transmitted along a stretched wire by means of transverse harmonic waves. The wave speed is 10 m/s, and the linear mass density of the wire is 0.01 kg/m. The power source oscillates with an amplitude of 0.50 mm. (a) What average power is transmitted along the wire if the frequency is 400 Hz? (b) The power transmitted can be increased by increasing the tension in the wire, the frequency of the source, or the amplitude of the waves. How would each of these quantities have to be changed to effect an increase in power by a factor of 100 if it is the only quantity changed? (c) Which of the quantities would probably be the easiest to change? (a) Apply Equ. 15-19 P = 1/2(0.01)(2π × 400 × 0.0005) 2(10) W

Chapter 15

Wave Motion = 0.079 W

(b) Since P is proportional to f 2, A2, and v, and v is proportional to F , the frequency or amplitude could be increased by a factor of 10, but the tension would have to be increased by a factor of 10,000. Evidently, it will be simplest to increase the amplitude to 5.0 mm. 33* ·

A sound wave in air produces a pressure variation given by

p(x,t) = 0.75 cos

π (x − 340t) 2

where p is in pascals, x is in meters, and t is in seconds. Find (a) the pressure amplitude of the sound wave, (b) the wavelength, (c) the frequency, and (d) the speed? (a) p 0 = 0.75 Pa. (b) From Problem 24(d), we see that 2π /λ = π /2, λ = 4 m. (c) f = v/λ = 85 Hz. (d) v = 340 m/s. 34 · (a) Middle C on the musical scale has a frequency of 262 Hz. What is the wavelength of this note in air? (b) The frequency of the C an octave above middle C is twice that of middle C. What is the wavelength of this note in air? (a) λ = v/f = 340/262 m = 1.30 m. (b) λ′ = λ/2 = 0.65 m. 35 · (a) What is the displacement amplitude for a sound wave having a frequency of 100 Hz and a pressure amplitude of 10−4 atm? (b) The displacement amplitude of a sound wave of frequency 300 Hz is 10−7 m. What is the pressure amplitude of this wave? (a) From Equ. 15-22, s0 = p 0/(ρωv) s0 = [10.1/(1.29 × 2π × 100 × 340)] m = 3.67 × 10−5 m (b) p 0 = s0ρωv p 0 = 1.29 × 300 × 2π × 340 × 10−7 Pa = 8.27 × 10−2 Pa 36 · (a) Find the displacement amplitude of a sound wave of frequency 500 Hz at the pain-threshold pressure amplitude of 29 Pa. (b) Find the displacement amplitude of a sound wave with the same pressure amplitude but a frequency of 1 kHz. (a) See Problem 35. s0 = 3.67 × 10−5 × 29/(10.1 × 5) m = 2.11 × 10−5 m. (b) s0 = (2.11 × 10−5/2) m = 1.05 × 10−5 m. 37* · A typical loud sound wave with a frequency of 1 kHz has a pressure amplitude of about 10−4 atm. (a) At t = 0, the pressure is a maximum at some point x1. What is the displacement at that point at t = 0? (b) What is the maximum value of the displacement at any time and place? (Take the density of air to be 1.29 kg/m3.) (a) When p is a maximum, s = 0. (b) s0 = p 0/ρωv = 3.67 × 10−6 m. 38 · (a) Find the displacement amplitude of a sound wave of frequency 500 Hz at the threshold-of-hearing pressure amplitude of 2.9 × 10−5 Pa. (b) Find the displacement amplitude of a wave of the same pressure amplitude but a frequency of 1 kHz. From Problem 36 it follows that (a) s0 = 2.11 × 10−11 m; (b) s0 = 1.05 × 10−11 m. 39 · A piston at one end of a long tube filled with air at room temperature and normal pressure oscillates with a frequency of 500 Hz and an amplitude of 0.1 mm. The area of the piston is 100 cm2. (a) What is the pressure amplitude of the sound waves generated in the tube? (b) What is the intensity of the waves? (c) What average power is required to keep the piston oscillating (neglecting friction)? (a) p 0 = s0ρωv p 0 = (10−4 × 1.29 × 2π × 500 × 340) Pa = 138 Pa

Chapter 15

306

Wave Motion

(b) I = 1/2ρω2s02v = 1/2p 02/ρ v (c) Pav = IA; A = 10−2 m2

I = 1/2(1382/1.29 × 340) W/m2 = 21.7 W/m2 Pav = 0.217 W

40 · A spherical source radiates sound uniformly in all directions. At a distance of 10 m, the sound intensity level is 10−4 W/m2. (a) At what distance from the source is the intensity 10−6 W/m2? (b) What power is radiated by this source? (a) I ∝ 1/r2; r(I = 10−6) = 100 m. (b) P = IA = 4π r2I = 0.126 W. 41* · A loudspeaker at a rock concert generates 10−2 W/m2 at 20 m at a frequency of 1 kHz. Assume that the speaker spreads its energy uniformly in three dimensions. (a) What is the total acoustic power output of the speaker? (b) At what distance will the intensity be at the pain threshold of 1 W/m2? (c) What is the intensity at 30 m? (a) P = 4π r2I = 4π × 400 × 10−2 W = 50.3 W. (b) Since I ∝ 1/r2, r at pain threshold is 2.0 m. (c) I = 10−2(4/9) = 4.44 × 10−3 W/m2. 42 ·· When a pin of mass 0.1 g is dropped from a height of 1 m, 0.05% of its energy is converted into a sound pulse with a duration of 0.1 s. (a) Estimate the range at which the dropped pin can be heard if the minimum audible intensity is 1011 W/m2. (b) Your result in (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 10−8 W/m2 for the sound to be heard, estimate the range at which the dropped pin can be heard. (In both parts, assume that the intensity is P/4π r2.) (a) Sound energy is 5 × 10−4(mgh) = 5 × 10−4 × 1 × 10−4 × 9.81 J = 4.9 × 10−7 J; Pav = E/∆t = 4.9 × 10−6 W = 4π r2 × 10−11 W; so r ≈ 200 m. (b) r ≈ 200/ 1000 = 6.24 m. 43 · True or false: A 60-dB sound has twice the intensity of a 30-dB sound. False 44 · What is the intensity level in decibels of a sound wave of intensity (a) 10−10 W/m2, and (b) 10−2 W/m2? Use Equ. 15-29. (a) β = 20 dB. (b) β = 100 dB. 45* · Find the intensity of a sound wave if (a) β = 10 dB, and (b) β = 3 dB. (c) Find the pressure amplitudes of sound waves in air for each of these intensities. (a), (b) Use Equ. 15-29 (a) I = 10−11 W/m2; (b) I = 2 × 10−12 W/m2 (a) p 0 = 9.37 × 10−5 Pa; (b) p 0 = 4.19 × 10−5 Pa (c) p = 2 Iρv 0

46 · The sound level of a dog's bark is 50 dB. The intensity of a rock concert is 10,000 times that of the dog's bark. What is the sound level of the rock concert? β = βdog + 40 dB = 90 dB. 47 · Two sounds differ by 30 dB. The intensity of the louder sound is IL and that of the softer sound is IS. The value of the ratio IL /IS is (a) 1000. (b) 30. (c) 9. (d) 100. (e) 300. (a) is correct. 48 ·

Show that if the intensity is doubled, the intensity level increases by 3.0 dB. ∆β = 10 log 2 = 3.01 ≈ 3.0.

49*· What fraction of the acoustic power of a noise would have to be eliminated to lower its sound intensity level from 90 to 70 dB?

Chapter 15

Wave Motion

99% must be eliminated so that the power is reduced by factor of 100. 50 ·· Normal human speech has a sound intensity level of about 65 dB at 1 m. Estimate the power of human speech. Determine I at 1 m. I = 10−12 × 106.5 W/m2 = 10−5.5 W/m2 = 3.16 × 10−6 W/m2 = P/4π . P = 4 × 10−5 W. 51 ·· A spherical source radiates sound uniformly in all directions. At a distance of 10 m, the sound intensity level is 80 dB. (a) At what distance from the source is the intensity level 60 dB? (b) What power is radiated by this source? (a) I(r) = I(10)/100; I ∝ 1/r2 r = 100 m (b) P = IA P = 10−4 × 4π × 102 W = 0.126 W 52 ·· A spherical source of intensity I0 radiates sound uniformly in all directions. Its intensity level is β1 at a distance r1, and β2 at a distance r2. Find β2 /β1.

β 2 log ( I 2 / I 0) log I 2 log [( r 12 / r 22) I 1] log I 1 + 2 log ( r 1 / r 2) β 1 + 20 log ( r1 / r 2) = = = = = . log I 1 log I 1 β 1 log ( I 1 / I 0) log I 1 β1 53* ·· A loudspeaker at a rock concert generates 10−2 W/m2 at 20 m at a frequency of 1 kHz. Assume that the speaker spreads its energy uniformly in all directions. (a) What is the intensity level at 20 m? (b) What is the total acoustic power output of the speaker? (c) At what distance will the intensity level be at the pain threshold of 120 dB? (d) What is the intensity level at 30 m? (a) β = 100 dB at 20 m. (b), (c), (d) See Problem 41; (d) 4.44 × 10−3 W/m2 = 96.5 dB. 54 ·· An article on noise pollution claims that sound intensity levels in large cities have been increasing by about 1 dB annually. (a) To what percentage increase in intensity does this correspond? Does this increase seem reasonable? (b) In about how many years will the intensity of sound double if it increases at 1 dB annually? (a) I/I0 = 100.1 = 1.26; increase in intensity is 26% annually. This is not reasonable; if true, the intensity level would increase by a factor of 10 in ten years. (b) To double, ∆β = 3, so the intensity level will double in 3 years. 55 ·· Three noise sources produce intensity levels of 70, 73, and 80 dB when acting separately. When the sources act together, their intensities add. (a) Find the sound intensity level in decibels when the three sources act at the same time. (b) Discuss the effectiveness of eliminating the two least intense sources in reducing the intensity level of the noise. (a) I1 = 10−5 W/m2, I2 = 2 × 10−5 W/m2, I3 = 10−4 W/m2. So Itot = 1.3 × 10−4 W/m2 ; βtot = 81.14 dB. (b) Eliminating the two least intense sources does not reduce the intensity level significantly. 56 ·· The equation I = Pav / 4π r2 is predicated on the assumption that the transmitting medium does not absorb any energy. It is known that absorption of sound by dry air results in a decrease of intensity of approximately 8 dB/km. The intensity of sound at a distance of 120 m from a jet engine is 130 dB. Find the intensity at 2.4 km from the jet engine (a) assuming no absorption of sound by air, and (b) assuming a diminution of 8 dB/km. (Assume that the sound radiates uniformly in all directions.) (a) ∆β = 20 log (r2/r1) 20 log (20) = 26; β(2.4 km) = (130 - 26) dB = 104 dB (b) Subtract 2.28 × 8 dB from result of (a) β = (104 - 18.2) dB = 85.8 dB

308

Chapter 15

Wave Motion

57* ··· Everyone at a party is talking equally loudly. If only one person were talking, the sound level would be 72 dB. Find the sound level when all 38 people are talking. Itot = 38I1; βtot = [10 log (38) + 72] dB = (15.8 + 72) dB = 88.8 dB. 58 ··· When a violinist pulls the bow across a string, the force with which the bow is pulled is fairly small, about 0.6 N. Suppose the bow travels across the A string, which vibrates at 440 Hz, at 0.5 m/s. A listener 35 m from the performer hears a sound of 60 dB intensity. With what efficiency is the mechanical energy of bowing converted to sound energy? (Assume that the sound radiates uniformly in all directions.) 1. Find the power delivered by the bow to the string Pin = Fv = 0.6 × 0.5 W = 0.3 W 2. Find the power of the sound emitted Pout = IA = (10−6 × 4π × 352) W = 0.0154 W 3. Efficiency = Pout/Pin η = 5.13% 59 ··· The noise le vel in an empty examination hall is 40 dB. When 100 students are writing an exam, the sounds of heavy breathing and pens traveling rapidly over paper cause the noise level to rise to 60 dB (not counting the occasional groans). Assuming that each student contributes an equal amount of noise power, find the noise level to the nearest decibel when 50 students have left. −6 −8 2 −6 2 1. Find the noise level increase due to one student ∆I100 = (10 - 10 ) W/m ≈ 10 W/m ; −8 2 ∆I1 = 10 W/m 2. Find I50 and β50 I50 = 5.1 × 10-7 W/m2; β50 = 57 dB 60 · If the source and receiver are at rest relative to each other but the wave medium is moving relative to them, will there be any Doppler shift in frequency? No 61* · The frequency of a car horn is f 0. What frequency is observed if both the car and the observer are at rest, but a wind blows toward the observer? (a) f 0 (b) Greater than f 0 (c) Less than f 0 (d) It could be either greater or less than f 0 . (e) It could be f 0 or greater than f 0, depending on how wind speed compares to speed of sound. (a) There is no relative motion of the source and receiver. 62 ·· Stars often occur in pairs revolving around their common center of mass. If one of the stars is a black hole, it is invisible. Explain how the existence of such a black hole might be inferred from the light observed from the other, visible star. The light from the companion star will be shifted about its mean frequency periodically due to the relative approach to and recession from the earth of the companion star as it revolves about the black hole. 63 · A conveyor belt moves to the right with a speed v = 300 m/min. A very fast piemaker puts pies on the belt at a rate of 20 per minute, and they are received at the other end by a pie eater. (a) If the piemaker is stationary, find the spacing λ between the pies and the frequency f with which they are received by the stationary pie eater. (b) The piemaker now walks with a speed of 30 m/min toward the receiver while continuing to put pies on the belt at 20 per minute. Find the spacing of the pies and the frequency with which they are received by the stationary pie eater. (c)

Chapter 15

Wave Motion

Repeat your calculations for a stationary piemaker and a pie eater who moves toward the piemaker at 30 m/min. (a) Since 20 pies per minute are placed on the belt, 20 pies per minute will be received. The time between the placing of the pies is 0.05 min, and during that time the belt moves 300 × 0.05 m = 15 m. Consequently, the spacing between the pies is λ = 15 m. (b) Relative to the piemaker, the belt moves at 270 m/min. Consequently, λ = 270 × 0.05 m = 13.5 m. Since the pies are traveling toward the receiver at 300 m/min, the number received per minute, i.e., the frequency, is given by f = 300/13.5 min−1 = 22.2 min−1. (c) If the receiver moves toward the piemaker, the spacing of the pies on the belt is, as in (a), 15 m. However, the speed of the belt relative to the receiver is 330 m/min, so the frequency f = 330/15 min−1 = 22 min−1. 64 · For the situation described in Problem 63, derive general expressions for the spacing of the pies λ and the frequency f with which they are received by the pie eater in terms of the speed of the belt v, the speed of the sender u s, the speed of the receiver u r, and the frequency f 0 with which the piemaker places pies on the belt. Equations 15-31 and 15-35 apply. Since the source is moving in the direction of the belt and the receiver is moving toward the observer, we must use the negative sign in the numerator of Equ. 15-31 and in the denominator of Equ. 1535.

Thus

λ’ =

v - us ; f0

f ’= f 0

v + ur . v - us

In Problems 65 through 70, a source emits sounds of frequency 200 Hz that travel through still air at 340 m/s. 65* · The sound source described above moves with a speed of 80 m/s relative to still air toward a stationary listener. (a) Find the wavelength of the sound between the source and the listener. (b) Find the frequency heard by the listener. (a) Apply Equ. 15-31 λ′ = (260/200) m = 1.3 m (b) Apply Equ. 15-35 f ′ = 200(340/260) Hz = 262 Hz 66 · Consider the situation in Problem 65 from the reference frame in which the source is at rest. In this frame, the listener moves toward the source with a speed of 80 m/s, and there is a wind blowing at 80 m/s from the listener to the source. (a) What is the speed of the sound from the source to the listener in this frame? (b) Find the wavelength of the sound between the source and the listener. (c) Find the frequency heard by the listener. (a) In the reference frame of the source, v = (340 - 80) m/s = 260 m/s. (b) Since f is unchanged, λ = 260/f = 1.3 m. (c) In the moving reference frame, the observer is approaching the source at 80 m/s. Consequently, applying Equ. 1535, we find f r = [200(1 + 80/260)] Hz = 262 Hz. 67 · The source moves away from the stationary listener at 80 m/s. (a) Find the wavelength of the sound waves between the source and the listener. (b) Find the frequency heard by the listener. Proceed as in Problem 65, changing signs as appropriate. (a) λ = 420/200 m = 2.1 m. (b) f = 200(340/420) Hz = 162 Hz. 68 ·

The listener moves at 80 m/s relative to still air toward the stationary source. (a) What is the wavelength of the

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Chapter 15

Wave Motion

sound between the source and the listener? (b) What is the frequency heard by the listener? (a) λ is unaffected by the motion of the observer λ = 340/200 m = 1.7 m (b) Apply Equ. 15-35 f ′ = [200(1 + 80/340)] Hz = 247 Hz 69* · Consider the situation in Problem 68 in a reference frame in which the listener is at rest. (a) What is the wind velocity in this frame? (b) What is the speed of the sound from the source to the listener in this frame, that is, relative to the listener? (c) Find the wavelength of the sound between the source and the listener in this frame. (d) Find the frequency heard by the listener. (a) Moving at 80 m/s in still air, the observer experiences a wind of 80 m/s. (b) Using the standard Galilean transformation, v′ = v + ur = 420 m/s. (c) The distance between wave crests is unchanged, so λ′ = λ = 1.7 m. (d) f ′ = v′/λ′ = 247 Hz. 70 · The listener moves at 80 m/s relative to the still air away from the stationary source. Find the frequency heard by the listener. Apply Equ. 15-35 f ′ = [200(1 - 80/340)] Hz = 152 Hz 71 · A jet is traveling at Mach 2.5 at an altitude of 5000 m. (a) What is the angle that the shock wave makes with the track of the jet? (Assume that the speed of sound at this altitude is still 340 m/s.) (b) Where is the jet when a person on the ground hears the shock wave? −1 o (a) u/v = 2.5; θ = sin−1(v/u) θ = sin (0.4) = 23.6

(b) From the diagram showing the wave front of the shock wave, it is evident that x = h/tan θ

x = (5000/tan 23.6o) m = 11.5 km

Chapter 15

Wave Motion

72 · If you are running at top speed toward a source of sound at 1000 Hz, estimate the frequency of the sound that you hear. Suppose that you can recognize a change in frequency of 3%. Can you use your sense of pitch to estimate your running speed? Top speed ≈ 10 m/s. ∆f/f 0 = u/v = 10/340 Hz = 0.0294 or 2.9% . No; it would be impossible to decide between speeds of 8 m/s and 10 m/s. 73* ·· A radar device emits microwaves with a frequency of 2.00 GHz. When the waves are reflected from a car moving directly away from the emitter, a frequency difference of 293 Hz is detected. Find the speed of the car. 1. The frequency f received by the car is given by Equ. 15-37a. 2. The car now acts as the source, sending signals of frequency f to the stationary radar receiver. 3. Consequently, f rec =

1 + v/c f = (1 + 2v/c) f 0 since v