A tuning-fork source of frequency 440 Hz moves at 10 m s^-1 directly towards a large stationary wall. A stationary observer standing behind the…

The source moves away from the observer, so the directly heard frequency is $f_1=\dfrac{fv}{v+v_s}=\dfrac{440\times340}{350}=427.43$ Hz. The wall is stationary and receives sound from the approaching source at $f_2=\dfrac{fv}{v-v_s}=\dfrac{440\times340}{330}=453.33$ Hz. A stationary reflector re-emits the same frequency, and the stationary observer hears the echo at $453.33$ Hz. The beat frequency is therefore $f_2-f_1=fv\left(\dfrac1{v-v_s}-\dfrac1{v+v_s}\right)=\dfrac{440\times340\times20}{340^2-10^2}=\dfrac{2\,992\,000}{115\,500}=25.90$ Hz. The trap is to treat the echo as another moving-source case with the observer receding; the wall is simply a stationary relay. Because both frequencies are heard simultaneously, the observer detects a periodic intensity variation at exactly this difference frequency.