The Doppler Effect — Question 1

Question 1

Equal closing speeds do not imply identical Doppler shifts

In still air with $c=340\,\mathrm{m/s}$, a source emits $f_s=500\,\mathrm{Hz}$. Compare (A) an observer moving toward a stationary source at $34\,\mathrm{m/s}$ and (B) a source moving toward a stationary observer at the same speed. All motion is along the line joining them.

1. Derive and calculate the observed frequency in each case.

2. Explain why equal source-observer closing speed gives different results, and compare their first-order small-speed expansions.

Question 1 - Solution

Equal closing speeds do not imply identical Doppler shifts

(a) Method and calculation

In A, stationary-source wavelength is $\lambda=c/f_s$, and the observer encounters fronts at speed $c+v$:$$f_A=f_s(1+v/c)=550\,\mathrm{Hz}.$$In B, successive fronts are separated by $(c-v)/f_s$, so$$f_B=\frac{cf_s}{c-v}=555.556\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The moving observer changes the rate of encountering an existing wave train; the moving source changes the wavefront spacing in the medium. With $M=v/c$,$$f_A/f_s=1+M,\qquad f_B/f_s=(1-M)^{-1}=1+M+M^2+\cdots.$$They agree to first order and differ at higher orders. Acoustic motion is referenced to the medium.

Physical check

Both approaching configurations raise frequency; both reduce to $f_s$ when the speed vanishes.