H.C.V  Vol 1,Pg 358, Q89.

 

A source emmiting a sound of frequency f is placed at a large distance from an observer. The source starts moving towards the observer with an accelaration a. Find the frequency heard by the observer corresponding to the wave emmitted just after the souce starts. The speed of sound in the medium is v.

 

I just can't understand the question properly. Please help and if possible solve.

When we proceed to solve the problem we find that the apparent time-period and hence app. frequency of the sound emitted by the accelerated source is a function of time(t). However as it is given to find the app. freq. of the sound just after the source starts moving so, we just have to consider the first time-period of the sound wave, i.e. from t = 0 to t = T.

   Let the the source and the observer be initially separated by a large distance D. At t=0 the source emits a wave pulse (with wave velocity = v). This wave pulse takes time D/v to reach the observer. Moreover the source itself starts moving from t=0 with acceleration a. From t = 0 to t = T, the source has progressed by distance = 0.T + (1/2)aT². = (1/2)aT². After one time-period, i.e. at t=T the wave emits the second wave pulse which has to travel a decreased distance i.e. (D - 1/2 aT² ). So, time taken for the second wave to reach the observer = (D - 1/2 aT² )/v. So, total time taken when calculated from t=0 is {T + (D - 1/2 aT² )/v}. So, apparent time period = time difference between the arrival of the two wave pulses = T - (1/2v)aT² = T{1 - (1/2v)aT} = T{(2v - aT)/2v}

  So, apparent frequency = 2v / {T(2v - aT} = 2fv / (2v - aT) = 2fv / (2v - a/f) = 2vf² / (2fv - a). From clarity you may refer to the Doppler's effect (Obs stationary source moving) section in Concepts of Physics.