An observer is riding on a bicycle and moving towards a hill at $18\ \text{km h}^{-1}$. He hears a sound from a source at some distance behind him directly as well as after its reflection from the hill. If the original frequency of the sound as emitted by the source is $640\ \text{Hz}$ and the velocity of sound in air is $320\ \text{m s}^{-1}$, the beat frequency between the two sounds heard by the observer will be ______ Hz.
Numerical value type. Enter your answer.
Answer: 20
Observer speed $= 5\ \text{m s}^{-1}$.
Direct sound (observer moving away from the source): $f_1 = 640\times\dfrac{320 - 5}{320} = 630\ \text{Hz}$.
Reflected sound (the stationary hill reflects $640\ \text{Hz}$; observer moving towards it): $f_2 = 640\times\dfrac{320 + 5}{320} = 650\ \text{Hz}$.
Beat frequency $= 650 - 630 = 20\ \text{Hz}$.
Solution by Sreeraj P, M.Sc Physics