What is the speed of the source?

Question

A source of sound moves toward you at speed Vs and away from Jane, who is standing on the other side of it. You hear the sound at twice the frequency as Jan.

What is the speed of the source? Assume that the speed of sound is 340 m/s.

Express your answer using two significant figures.

Answer 1

For a receding source (Jane's view) the observed frequency is

f(observed) = [ V/(V-Vs)] f(source)

For an approaching source (your view) the observed frequency is

f(observed) = [ V/(V+Vs)] f(source)

Call the approaching observed frequency fo' and the receding observed frequency fo and f(source) = fs

The problem states that fo' = 2fo

fo = [ V/(V-Vs)] fs

2fo = [ V/(V+Vs)] fs

The second eq. becomes

fo = (1/2) [ V/(V+Vs)] fs

So equate the 1st and 2nd



[ V/(V-Vs)] fs = (1/2) [ V/(V+Vs)] fs

V/(V-Vs) =  V/2(V+Vs)

1/(340 -Vs) = 1/(680+2Vs)

340-Vs = 680 +2Vs

3Vs = -340

Vs = -113.333333

Vs= -110 m/s (in Doppler terms the minus sign means receding; so this is from jane's view)

Vs = 110 m/s

Answer 2

Lets denote wavelength of the sound  ÃƒÂŽÃ‚»,

frequency of sound f = c/λ,,

and distance between you and Jane D, and

s = 340 m/s.

The source travelled from Jane to you in time t = D/v.

During this time number of wave periods produced by the source was

N = f t = (s/λ) D/v

You registered this number of periods N plus those periods which were between you and Jane D/λ.

Jane registered this number of periods N less those periods which remain between you and Jane D/λ.

The equation:

N + D/λ = 2(N - D/λ)

N + D/λ = 2N - 2D/λ

N = 3 D/λ

(s/λ) D/v = 3D/λ

s/v = 3

v = 1/3 s