Chapter 17   [HOMEWORK 6]                      Return by Wednesday, December 1 

 

1 1(1.50 )sin [(0.900 ) (315 ) ].

Problem 1 [20 pts] a) The pressure in a traveling sound wave is given by the equation

p Pa m x s tπ − −Δ = −

( ) ( ) ( )( )s x, t 6.0 nm cos kx+ 3000 rad / s t+

Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and (d) speed of the wave. b) If the form of a sound wave traveling through air is

= ϕ , how much time does any given air molecule along the path take to move between displacements s = + 2.0 nm and s = - 2.0 nm? Solution: (a) The amplitude of a sinusoidal wave is the numerical coefficient of the sine (or cosine) function: pm = 1.50 Pa.

We identify k = 0.9π and ω = 315 π (in SI units), which leads to f = ω /2 π = 158 Hz.

We also obtain λ = 2π/k = 2.22 m.

The speed of the wave is v = λ /k = 350 m/s.

(b) Without loss of generality we take x = 0, and let t = 0 be when s = 0. This means the phase is φ = -π/2 and the function is s = (6.0 nm)sin(ωt) at x = 0. Noting that ω = 3000 rad/s, we note that at t = sin-1(1/3)/ω = 0.1133 ms the displacement is s = +2.0 nm. Doubling that time (so that we consider the excursion from –2.0 nm to +2.0 nm) we conclude that the time required is 2(0.1133 ms) = 0.23 ms.

Chapter 17   [HOMEWORK 6]                      Return by Wednesday, December 1 

 

Problem 2 [20 pts] Earthquakes generate sound waves inside Earth. Unlike a gas, Earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S waves is about 4.5 km/s, and that of P waves 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P waves arrive 3.0 min before the first S waves. If the waves travel in a straight line, how far away does the earthquake occur? Solution: If d is the distance from the location of the earthquake to the seismograph and vs is the speed of the S waves then the time for these waves to reach the seismograph is ts. = d/vs. Similarly, the time for P waves to reach the seismograph is tp = d/vp. The time delay is

Δt = (d/vs) – (d/vp) = d(vp – vs)/vsvp,

so

3(4.5 km/s)(8.0 km/s)(3.0 min)(60s /min) 1.9 10 km.( ) 8.0 km/s 4.5km/s

s p

p s

v v td

v vΔ

= = = ×− −

We note that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds.

 

Chapter 17   [HOMEWORK 6]                      Return by Wednesday, December 1 

 

Problem 3 [20 pts] A point source emits 30.0 W of sound isotropically. A small microphone intercepts the sound in an area of 0.750 cm2, 200 m from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone. Solution: (a) The intensity is

5 22 2

30.0W 5.97 10 W/m .4 (4 )(200m)

PIr

−= = = ×π π

5 2 2 4 2 2 90 (6.0 10 W/m )(0.750cm )(10 m / cm ) 4.48 10 W.P IA − − −′ = = = × = ×

(b) Let A (= 0.750 cm2) be the cross-sectional area of the microphone. Then the power intercepted by the microphone is

Chapter 17   [HOMEWORK 6]                      Return by Wednesday, December 1 

 

Problem 4 [20 pts] An ambulance with a siren emitting a whine at 1,600 Hz overtakes and passes a cyclist pedaling a bike at 2.44 m/s. After being passed, the cyclist hears a frequency of 1590 Hz. How fast is the ambulance moving?

Solution:

 We have v = 343 m/s, vD = 2.44 m/s,

f '= 1590 Hz and f = 1600 Hz. Thus,

( ) 4.61m/s.DS D

S

v v ff f v v v vv v f

⎛ ⎞+′ = ⇒ = + − =⎜ ⎟ ′+⎝ ⎠

Chapter 17   [HOMEWORK 6]                      Return by Wednesday, December 1 

 

Problem 5 [20 pts] A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 29.9 m/s, the speed of surface B is 65.8 m/s, and the speed of sound is 329 m/s. The source emits waves at frequency 1200 Hz as measured in the source frame. In the reflector frame, what are the (a) frequency and (b) wavelength of the arriving sound waves? In the source frame, what are the (c) frequency and (d) wavelength of the sound waves reflected back to the source?

Solution:

(a) vD = 65.8 m/s and vS = 29.9 m/s, and we choose signs so that f ' is larger than f:

3329 m/s 65.8 m/s 1.58 10 Hz.329 m/s 29.9 m/s

f f⎛ ⎞+′ = = ×⎜ ⎟−⎝ ⎠

(b) The wavelength is λ = v/f ' = 0.208 m.

(c) The wave (of frequency f ') “emitted” by the moving reflector (now treated as a “source,” so vS = 65.8 m/s) is returned to the detector (now treated as a detector, so vD = 29.9 m/s) and registered as a new frequency f '':

3329 m/s 29.9 m/s 2.16 10 Hz.329 m/s 65.8 m/s

f f⎛ ⎞+′′ ′= = ×⎜ ⎟−⎝ ⎠

/ ′′

(d) This has wavelength v f = 0.152 m.