A turning fork vibrating at 500Hz falls from rest accelerates at 10 m//s^(2). Velocity of the tuning fork when waves with a frequency of 475 Hz reach the release point is ( Take the speed of sound in air to be 340 m//s).

A turning fork vibrating at 500Hz falls from rest accelerates at 10 m/

1.	A turning fork vibrating at 500Hz falls from rest accelerates at 10 m//s^(2). Velocity of the tuning fork when waves with a frequency of 475 Hz reach the release point is ( Take the speed of sound in air to be 340 m//s).
Answer» `1.79 m//s` `17.9 m//s` `35.8 m//s` `3.58 m//s` Solution :Let` t'` be the time at which the tuning fork emits a sound wave which REACHES the release point at `(t - t')` The apparent frequency received at the release point `v' = (v_(0) v)/( v + v_(s))` `V' = 475 HZ , v_(0) = 500 Hz , v = 340 m//s` `475 = (500 xx 340)/(340 + v_(s)) rArrv_(s) = 1.79 m//s`