A tuning fork vibrating at 513 Hz falls from rest and accele

A tuning fork vibrating at 513 Hz falls from rest and accelerates at 9.80 m/s2. How far below the point of release is the tuning fork when waves of frequency of 481 Hz reach the release point? (Take the speed of sound in air to be 343 m/s.) m

Solution

Using Doppler Effect:

f = (v + Vr)f0 / (V + Vs)

where V = speed of wave

Vr = speed of receiver = 0

Vs = speed of source = v

481 = (343 x 513 ) / (343 + v)

343 + v = 365.81

v = 22.81 m/s

now using vf^2 -vi^2 = 2ad

22.81^2 - 0 = 2(9.8)(d)

d = 26.545 m