A source emitting sound of frequency 500 Hz moves at 20 m s^-1 directly towards a large flat vertical wall. A stationary observer stands on the line…

The source moves away from the observer, so the direct sound has $f_1=\dfrac{500\times340}{340+20}=472.2\ \mathrm{Hz}$ and wavelength $\lambda_1=\dfrac{340+20}{500}=0.72\ \mathrm{m}$ (waves are stretched behind the source). The wall receives sound from a source approaching it: wavelength in front of the source is $\lambda_2=\dfrac{340-20}{500}=0.64\ \mathrm{m}$, and the wall (stationary) reflects the same wavelength, so $f_2=340/0.64=531.25\ \mathrm{Hz}$ is heard by the observer, who is at rest. Beats: $f_2-f_1=59.0\ \mathrm{Hz}$. (A) true. (B) false, since $0.68\ \mathrm{m}$ is $v/f$ and the reflected wavelength is $0.64\ \mathrm{m}$. (C) false, $531$ Hz belongs to the reflected sound, while the direct is $472$ Hz. (D) false, the direct wavelength is $0.72\ \mathrm{m}$.