Chapter 15 Traveling WavesConceptual Problems1 A rope hangs vertically from the ceiling. A pulse is sent up the rope. Does the pulse travel faster, slower, or at a constant speed as it moves toward the ceiling? Explain your answer. Determine the Concept The speed of a transverse wave on a uniform rope increases with increasing tension. The waves on the rope move faster as they move toward the ceiling because the tension increases due to the weight of the rope below the pulse. 5 To keep all of the lengths of the treble strings (unwrapped steel wires) in a piano all about the same order of magnitude, wires of different linear mass densities are employed. Explain how this allows a piano manufacturer to use wires with lengths that are the same order of magnitude. Determine the Concept The resonant (standing wave) frequencies on a string are inversely proportional to the square root of the linear density of the string f = TT . Thus extremely high frequencies (which might otherwise require

(

)

very long strings) can be accommodated on relatively short strings if the strings are linearly denser that the high frequency strings. High frequencies are not a problem as they utilize short strings anyway. 11 At a given location, two harmonic sound waves have the same amplitude, but the frequency of sound A is twice the frequency of sound B. How do their average energy densities compare? (a) The average energy density of A is twice the average energy density of B. (b) The average energy density of A is four times the average energy density of B. (c) The average energy density of A is 16 times the average energy density of B. (d) You cannot compare the average energy densities from the data given. Determine the Concept The average energy density of a sound wave is given by 2 av = 1 2 s0 where is the average density of the medium, s0 is the 2 displacement amplitude of the molecules making up the medium, and is the angular frequency of the sound waves. Express the average energy density of sound A: The average energy density of sound B is given by:2 2 av, A = 1 A A s0, A 2

2 2 av, B = 1 B B s0, B 2

307

308

Chapter 152 2 av, A 1 A A s0, A 2 = 2 2 av, B 1 B B s0, B 2

Dividing the first of these equation by the second yields: Because the sound waves are identical except for their frequencies:

2 av, A A 2f A f A = = 2 = av, B B 2f B f B

2

2

Because fA = 2fB:

av, A 2 f B = 4 (b ) is correct. = av, B f B

2

19 Sound waves in air encounter a 1.0-m wide door into a classroom. Due to the effects of refraction, the sound of which frequency is least likely to be heard by all the students in the room, assuming the room is full? (a) 600 Hz, (b) 300 Hz, (c) 100 Hz, (d) All the sounds are equally likely to be heard in the room. (e) Diffraction depends on wavelength not frequency, so you cannot tell from the data given. Determine the Concept If the wavelength is large relative to the door, the diffraction effects are large and the waves spread out as they pass through the door. Because were interested in sounds that are least likely to be heard everywhere in the room, we want the wavelength to be short and the frequency to be high. Hence (a ) is correct. 21 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 by measuring the Doppler frequency shift of the light observed from the other, visible star. Determine the Concept The light from the visible star will be shifted about its mean frequency periodically due to the relative approach toward and recession away from Earth as the star revolves about the common center of mass.

Speed of Waves33 (a) Compute the derivative of the speed of a wave on a string with respect to the tension dv/dFT, and show that the differentials dv and dFT obey dv v = 1 dFT FT . (b) A wave moves with a speed of 300 m/s on a string that is 2

under a tension of 500 N. Using to the differential approximation, estimate how much the tension must be changed to increase the speed to 312 m/s. (c) Calculate FT exactly and compare it to the differential approximation result in Part (b). Assume that the string does not stretch with the increase in tension.

Traveling Waves

309

Picture the Problem (a) The speed of a transverse wave on a string is given by v = FT where FT is the tension in the wire and is its linear density. We can

differentiate this expression with respect to FT and then separate the variables to show that the differentials satisfy dv v = 1 dFT FT . (b) Well approximate the 2 differential quantities to determine by how much the tension must be changed to increase the speed of the wave to 312 m/s. (c) We can use v = FT to obtain an exact expression for FT, (a) Evaluate dv/dFT: dv d FT 1 1 1 v = = = dF dFT 2 FT 2 FTdv 1 dFT = v 2 FT

Separate the variables to obtain:

(b) Solve the equation derived in Part (a) for dFT:

dFT = 2 FT

dv v

Approximate dFT with FT and dv with v to obtain:

FT = 2 FT

v v

Substitute numerical values and evaluate FT:

312 m/s 300 m/s FT = 2(500 N ) 300 m/s = 40 N

(c) The exact value for (F)exact is given by:

(F )exact = FT,2 FT,1FT,1 FT,2

(1)

Express the wave speeds for the two tensions: Dividing the second equation by the first and simplifying yields:

v1 =

FT,2

and v2 =

v2 = v1

FT,1

=

FT,2 FT,1

FT,2

v = FF,1 2 v 1

2

310

Chapter 15

Substituting for FT,2 in equation (1) yields:

(FT )exact

v = FF,1 2 FT,1 v 1 v 2 = FF,1 2 1 v1

2

Substitute numerical values and evaluate (FT)exact:

(FT )exact

312 m/s 2 = (500 N ) 1 300 m/s = 40.8 N

The percent error between the exact and approximate values for FT is:

(FT )exact FT (FT )exact

=

40.8 N 40.0 N 40.8 N

2%

Harmonic Waves on a String39 A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 80 N, has an amplitude of 5.0 cm. Each point on the string moves with simple harmonic motion at a frequency of 10 Hz. What is the power carried by the wave propagating along the string? Picture the Problem The average power propagated along the string by a harmonic wave is Pav = 1 2 A2v, where v is the speed of the wave, and , , and 2

A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Express and evaluate the power propagated along the string: The speed of the wave on the string is given by: Substitute for v to obtain: Pav = 1 2 A2v 2

v=

FT

Pav = 1 2 A 2 2

FT

Traveling Waves Substitute numerical values and evaluate Pav:Pav =1 2

311

(4 )(0.050 kg/m )(10 s ) (0.050 m)21 2

2

80 N = 9 .9 W 0.05 kg/m

45 Power is to be transmitted along a taut string by means of transverse harmonic waves. The wave speed is 10 m/s and the linear mass density of the string is 0.010 kg/m. The power source oscillates with an amplitude of 0.50 mm. (a) What average power is transmitted along the string if the frequency is 400 Hz? (b) The power transmitted can be increased by increasing the tension in the string, the frequency of the source, or the amplitude of the waves. By how much would each of these quantities have to increase to cause an increase in power by a factor of 100 if it is the only quantity changed? Picture the Problem The average power propagated along a string by a harmonic wave is Pav = 1 2 A2 v where v is the speed of the wave, and , , and 2

A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively.

(a) Express the average power transmitted along the string:Substitute numerical values and evaluate Pav:

Pav = 1 2 A2 v = 2 2 f 2 A2 v 2

Pav = 2 2 (0.010 kg/m ) 400 s 1 0.50 10 3 = 79 mW

(

( ) m ) (10 m/s )2

2

(b) Because Pav f 2 , increasing the frequency by a factor of 10 would increase the power by a factor of 100. Because Pav A2 , increasing the amplitude by a factor of 10 would increase the power by a factor of 100. Because Pav v and v F , increasing the tension by a factor of 104 would increase v by a factor of 100 and the power by a factor of 100.

312

Chapter 15

Harmonic Sound Waves49 (a) What is the displacement amplitude for a sound wave with a frequency of 100 Hz and a pressure amplitude of 1.00 104 atm? (b) The displacement amplitude of a sound wave of frequency 300 Hz is 1.00 107 m. Assuming the density of air is 1.29 kg/m3, what is the pressure amplitude of this wave? Picture the Problem The pressure amplitude depends on the density of the medium , the angular frequency of the sound wave , the speed of the wave v, and the displacement amplitude s0 according to p0 = vs0 .

(a) Solve p0 = vs0 for s0:

s0 =

p0 v

Substitute numerical values and evaluate s0:s0 =

(1.00 10 atm)(1.01325 10 Pa/atm) = 3.64 10 2 (1.29 kg/m )(100 s )(343 m/s)4

5

5

3

1

m = 36.4 m

(b) Use p0 = vs0 to find p0:p0 = 2 1.29 kg/m 3 300 s 1 (343 m/s ) 1.00 10 7 m = 83.4 mPa

(

)(

)

(

)

Waves in Three Dimensions: Intensity55 A loudspeaker at a rock concert generates a sound that has an intensity level equal to 1.00 102 W/m2 at 20.0 m and has a frequency of 1.00 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 sound intensity be at the pain threshold of 1.00 W/m2? (c) What is the sound intensity at 30.0 m? Picture the Problem Because the power radiated by the loudspeaker is the product of the intensity of the sound and the area over which it is distributed, we can use this relationship to find the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power.

(a) Use Pav = 4r 2 I to find the total acoustic power output of the speaker:

Pav = 4 (20.0 m ) 1.00 10 2 W/m 22

(

)

= 50.27 W = 50.3 W

Traveling Waves (b) Relate the intensity of the sound at 20 m to the distance from the speaker: Relate the threshold-of-pain intensity to the distance from the speaker: Divide the first of these equations by the second and solve for r: (c) Use I = at 30.0 m:Pav to find the intensity 4r 21.00 10 2 W/m 2 = 4 (20.0 m ) Pav

313

2

1.00 W/m 2 =

Pav 4r 2

r=

(1.00 10 )(20.0 m )2

2

= 2.00 m

I (30.0 m ) =

50.3 W 2 4 (30.0 m )

= 4.45 10 3 W/m 2

*Intensity Level57 What is the intensity level in decibels of a sound wave of intensity (a) 1.00 1010 W/m2 and (b) 1.00 102 W/m2? Picture the Problem The intensity level of a sound wave, measured in decibels, is given by = (10 dB) log(I I 0 ) where I0 = 1012 W/m2 is defined to be

the threshold of hearing. (a) Using its definition, calculate the intensity level of a sound wave whose intensity is 1.00 1010 W/m2: 1.00 10 10 W/m 2 = (10 dB)log 10 12 W/m 2 = 10 log10 2 = 20.0 dB

(b) Proceed as in (a) with I = 1.00 102 W/m2:

= (10 dB)log

1.00 10 2 W/m 2 2 12 10 W/m

= 10 log1010 = 100 dB

67 The noise intensity level at some location in an empty examination hall is 40 dB. When 100 students are writing an exam, the noise level at that location increases to 60 dB. Assuming that the noise produced by each student contributes an equal amount of acoustic power, find the noise intensity level at that location when 50 students have left.

314

Chapter 15

Picture the Problem Because the sound intensities are additive, well find the noise intensity level due to one student by subtracting the background noise intensity from the intensity due to the students and dividing by 100. Then, well use this result to calculate the intensity level due to 50 students.

Express the intensity level due to 50 students: Find the sound intensity when 100 students are writing the exam:

50 = (10 dB) log

50 I1 I0

I 60 dB = (10 dB) log 100 I 0 and I100 = 10 6 I 0 = 10 6 W/m 2 I background 40 dB = (10 dB) log I 0 and I background = 10 4 I 0 = 10 8 W/m 2

Find the sound intensity due to the background noise:

Express the sound intensity due to the 100 students: Find the sound intensity due to 1 student: Substitute numerical values and evaluate the noise intensity level due to 50 students:

I100 I background = 10 6 W/m 2 10 8 W/m 2 10 6 W/m 2 I 100 I background = 10 8 W/m 2

100

50

50(1.00 10 8 W/m 2 ) = (10 dB)log 10 12 W/m 2= 57 dB

String Waves Experiencing Speed Changes69 Consider a taut string with a mass per unit length 1, carrying transverse wave pulses that are incident upon a point where the string connects to a second string with a mass per unit length 2. (a) Show that if 2 = 1, then the reflection coefficient r equals zero and the transmission coefficient equals +1. (b) Show that if 2 >> 1, then r 1 and 0; and (c) if 2 > 1 then v1 >> v2:

and

=

2v2 = v2 + v1

2 0 v 1+ 1 v2 1

(c) If 2 > v1:

v1 v v v2 r= 2 1 = 1 v1 v2 + v1 1+ v2

and

=

2v2 = v2 + v1

2 2 v1 1+ v2

The Doppler Effect79 The Doppler effect is routinely used to measure the speed of winds in storm systems. As the manager of a weather monitoring station in the Midwest, you are using a Doppler radar system that has a frequency of 625 MHz to bounce a radar pulse off of the raindrops in a swirling thunderstorm system 50 km away. You measure the reflected radar pulse to be up-shifted in frequency by 325 Hz. Assuming the wind is headed directly toward you, how fast are the winds in the storm system moving? Hint: The radar system can only measure the component of the wind velocity along its line of sight. Picture the Problem Because the wind is moving toward the weather station (radar device), the frequency fr the raindrops receive will be greater than the frequency emitted by the radar device. The radar waves reflected from the raindrops, moving toward the stationary detector at the weather station, will be of a still higher frequency fr. We can use the Doppler shift equations to derive an expression for the radial speed u of the wind in terms of difference of these frequencies.

Use Equation 15-41a to express the frequency fr received by the raindrops in terms of fs, ur, and c: The waves reflected by the drops are like waves re-emitted by a source moving toward the source at the weather station:

fr =

c ur c + ur fs = fs c us c

(1)

f r' =

c c ur fr = c u fr c us s

(2)

Traveling Waves Substitute equation (1) in equation (2) to eliminate fr:

317

c c + ur c + ur f r' = c u c fs = c u fs s s ur 1 1+ c f = 1 + u r 1 u r f = s s c c 1 ur c u u f r' 1 + r 1 + r f s c c u = 1 + r f s c 2

u u Because ur