A tuning fork vibrating at 512 Hz falls from rest and accele

A tuning fork vibrating at 512 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tuning fork when waves of frequency 485 Hz reach the release point?

Solution

This the problem of doppler effect.

The formula is: f=f0*((v+vr)/(v+vs))

f-observed frequency

f0-actual emitted frequency

v-velocity of the waves in the medium

vr-velocity of the receiver relative to the medium

vs-velocity of the source relative to the medium

Here vr=0

therefore our formula becomes,

f=f0*(v/(v-vs)) (since the source is moving away from its release point)

485=512*(343/(343-vs))

vs=-19.09m/s

the time in which the tuning fork attains the velocity 19.09m/s:

t1=vs/a (a-acceleration of the tuning fork)

t1=19.09/9.8

=1.95m/s

the distance travelled by the tuning fork in time t1 is,

h=1/2*a*t^2

=1/2*9.8*(1.95)^2

=18.6m.