Question

A source emitting a sound of frequency v is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v.

Answer 1

Frequency Heard is $\frac {2Vv^2}{2Vv-a}$

Explanation:

• The first pulse emitted (at t = 0 s) by the source at A  

reaches the observer at O in time $\frac {AO}{V} =\frac {d}{V}$, where AO = d.  

 

• Let the second pulse be emitted by the source at time t = T.

In this time the source moves a distance  

s = $0\times T+ \frac {1}{2}aT^2 = \frac {1}{2}aT^2$

 

• Now the source is at a distance = d - s

= d - $\frac {1}{2}aT^2$

The pulse produced here will reach the listener in time = $\frac {d - \frac {1}{2}aT^2}{V}$

• Hence the time between two consecutive pulses heard  

= ${T+ \frac {d - \frac {1}{2}aT^2}{V}} - \frac {d}{V}$

= $T - \frac {aT^2}{2V}$

= Time period of the pulse heard = T'  

• Hence the frequency of the sound heard  

v' = $\frac {1}{T'}$

= $\frac {1}{T - \frac {aT^2}{2V}}$

But T = $\frac {1}{v}$  

v' = $\frac {1}{\frac {1}{v} - \frac {a}{2Vv^2}}$

= $\frac {2Vv^2}{2Vv-a}$