The Doppler Effect — Question 7

Question 7

The source angle belongs to the emission event

A source moves with constant velocity $\mathbf v_s$ through still air. A stationary observer receives sound emitted at time $t_e$ from distance $R(t_e)$. Let $\hat{\mathbf n}$ point from that emission position to the observer. Arrival time is $t_r=t_e+R(t_e)/c$.

1. Derive the received frequency in terms of $\hat{\mathbf n}\cdot\mathbf v_s$.

2. For $f_s=600\,\mathrm{Hz}$, $c=340\,\mathrm{m/s}$, $v_s=100\,\mathrm{m/s}$ and $\hat{\mathbf n}\cdot\mathbf v_s=0.60v_s$, find $f_o$. What happens when the projection is zero?

Question 7 - Solution

The source angle belongs to the emission event

(a) Method and calculation

For a fixed observer, $dR/dt_e=-\hat{\mathbf n}\cdot\mathbf v_s$. Thus$$\frac{dt_r}{dt_e}=1-\frac{\hat{\mathbf n}\cdot\mathbf v_s}{c}.$$The same emitted cycle count occupies a different arrival interval, giving$$f_o=f_s\frac{dt_e}{dt_r}=\frac{f_s}{1-\hat{\mathbf n}\cdot\mathbf v_s/c}.$$

(b) Interpretation and further calculation

The projected velocity is $60\,\mathrm{m/s}$, so$$f_o=\frac{600}{1-60/340}=728.571\,\mathrm{Hz}.$$If the emission-direction projection is zero, $f_o=f_s$ in classical acoustics. Using the direction to the source at reception would generally select the wrong angle because the source has moved during propagation.

Physical check

The collinear approaching limit recovers $f_sc/(c-v_s)$. The derivation assumes a subsonic source so arrival order is regular.