Question

A source of sound emits waves at a frequency f= 650 Hz. An observer is located at a distance d= 170 m from the source. Use u=340 m/s for the speed of sound.

(a) Assume completely still air. How many wavefronts (full waves) N are there between the source and the observer?

N=

unanswered
(b) If the observer is moving away from the source at a (radial) velocity v=40 m/s, how does the number of wavefronts N found in part (a) change with time? For the answer, give the rate of change of N, namely dNdt (in Hz)

dN/dt=

unanswered
(c) By comparing the difference of the rate of wavefronts leaving and wavefronts entering the region between source and observer, calculate the frequency f′ observed by the moving observer. (in Hz)

hint: how does the difference relate to the rate of change of N you calculated in (b)?

f′=

unanswered
(d) Let us now assume that both source and observer are at rest, but wind blows at a constant speed v=20 m/s in the direction source towards observer. By comparing the difference of the rate of wavefronts leaving and wavefronts entering the region between source and observer, calculate the observed frequency f′? (in Hz)

f′=

Answer 1

d : distance
f : freq
v_s: 340m/s

a) N = df/(v_sound) (1)

b) asume that d = vt .
plug into eq(1):
N(t) = vtf/v_s => N(t) = (vf/v_s)t
and so d/dt of that is vf/v_s
prove this to yourself by working out the answer to (a) using this.

c) use ((v_s+v_0)/v_s)*f watch your signs, remember that if the source and observer are moving apart, f should be smaller. so it could be either (v_s+v_0) or (v_s-v_0) in the numerator.

Answer 2

d) f' = f

Answer 3

Anonymous can u explain c onemore time?

Answer 4

are you sure about d???
I thought the wind will increase the frequency of the source

Andy, for C)

f'=f(1- vobs/vsound)

Answer 5

Can some explain Part d of doppler shift. I used the following: f'= f*(1+speed of observer/speed of sound)
f'=450*(1+20/340)=450*1.0588=476.46. This is my last chance. The wind blows toward the observer. Any ideas. Thanks.

Answer 6

it doesnt matter if the wind blows, so its the same frequency. In your case 450.

Answer 7

guys where did you find those equations, lecture num or book chapt pls?

Answer 8

part d- f'=f
explanation: in case of wind blowing, we'll add the speed of wind to the speed of sound (vectorially). Hence in this case it will add up. Now since the both the observer and the source are stationary, there will be no change in the frequency of sound.
NOTE: frequency will change if there is relative motion between the observer and the source.
FORMULA: f'=f((v+wind-u1)/(v+wind-u2))

v-speed of sound
u1-observer speed
u2-source speed
wind-wind speed (add vectorially)