A student holds a tuning fork ocillating at 2456 Hz. He walks towards a wall at a constant speed of `1.33(m)/(s)`. (a) What beat frequency does the student observes between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of `5.00Hz`?

Explanation

An electronic burglar detector may listen for beasts between the signal it broadcasts and the Doppler shifted reflection from a moving surface. The experiment described here is more fun. The student hears beats at a rate that is controlled through his walking speed. We must thinks of a double Doppler shift, for the wave reaching the wall from the moving tuning fork and the wave reaching the moving auditor.
For this echo,
`f_("received back")=r_("radianted")(v+v_e)/(v-v_e)`
where `v_e` is the magnitude of the experimenter's velocity the beat frequency is
`f_b=|f_("received back")-f_("radiated")|`
Substituting gives
`f_b=f_("radiated")(2v_e)/(v-v_e)`
When approaching the wall
a. `f_b=(256Hz)(2(1.33(m)/(s)))/(343(m)/(s)-1.33(m)/(s))=1.99Hz`
b. When moving away from wall `v_e` appears with opposite signs, Solving for `v_e` gives
`v_e=f_b(v)/(2f_("radiated")-f_b)`
`=(5.00Hz)[(343(m)/(s))/(2(256Hz)-5.00Hz)]=3.38(m)/(s)`