A sound source S , emitting a sound of frequency 400 Hz and a receiver R of mass m are at the same point. R is performing SHM with the help of a spring of force constant K . At a time t = 0 , R is at the mean position and moving towards the right, as shown. At the same time, the source starts moving away from R with some acceleration a . The frequency registered by the receiver at a time t = 10 s is 250 Hz . What is

Step-by-step solution

The time period of oscillation T = 2 π m K = 10   s So at t = 10 s, the receiver passes through mean position towards the right with a speed v R = A ω = 100 π × π 5 = 20   m   s - 1 Using doppler’s formula f app = f 0 v - v R v + v S 250 = 400   320 - 20 320 + v S   ⇒   v S = 160   m   s - 1 So the velocity of the source is v S = 160   m   s - 1 at the time of emission of the wave which was received by the receiver at t = 10 &