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.

Options

A. `(vf^2)/(2vf-a)`
B. `(2vf^2)/(2vf+a)`
C. `(2vf^2)/(3vf-a)`
D. `(2vf^2)/(2vf-a)`

Answer

D

Explanation

Suppose at `t=0` distance between source and observer is I, First, wave pulse (say`p_1`) is emitted at this instant. This pulse will reach the observer after a time
`t_1=(l)/(v)` ..(i)
Source will emit the next pulse (say `p_2`) after a time
`T=(1)/(f)`.
During this time the source will move a distance `((1)/(2))aT^2` towards the observer. This pulse `p_2` will reach the observer in a time
`t_2=T+(l-(l)/(2)aT^2)/(v)` .(ii)
changed time period as observed by the observer is
`T'=t_2-t_1=T+(l)/(v)-(1)/(2)(aT^2)/(v)-(l)/(v)`
Substituting `T'=(1)/(f)` and `T=(1)/(f)` in the above equation, we get
`f'=(2vf^2)/(2vf-a)`