The Doppler Effect — Question 3

Question 3

Common translation changes wavelength but cancels the frequency shift

A source and observer move right together at $u=50\,\mathrm{m/s}$ through still air with $c=340\,\mathrm{m/s}$. Their separation is fixed, the observer is ahead, and the source emits $400\,\mathrm{Hz}$.

1. Find the wavefront spacing in the air and the observed frequency.

2. Compare the wavelength with that of a stationary source. Why does constant separation alone not explain every acoustic Doppler experiment?

Question 3 - Solution

Common translation changes wavelength but cancels the frequency shift

(a) Method and calculation

Forward spacing in the air is$$\lambda_+=\frac{c-u}{f_s}=\frac{290}{400}=0.725\,\mathrm m.$$The observer moves with the source and encounters these fronts at $c-u$, so$$f_o=\frac{290}{0.725}=400\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

A stationary source would give $\lambda_0=c/f_s=0.850\,\mathrm m$. Thus wavelength changes despite zero frequency shift between the translating endpoints. Cancellation follows from identical velocities relative to the same uniform medium:$$f_o/f_s=(c-u)/(c-u)=1.$$The medium remains essential for determining spacing and propagation; a distance-only rule is insufficient in general.

Physical check

This is classical, uniform translation through a nondispersive medium; it does not assert that the air itself moves with the apparatus.