A fixed source of sound emitting a certain frequency appears as `f_(a)` when the observer is approaching the source with `upsilon_(0)` and `f_(r)` when the observer recedes from the source with the same speed. The frequency of source is

Options

A. `(f_(r) + f_(a))/(2)`
B. `(f_(r) - f_(a))/(2)`
C. `sqrt(f_(a) f_(r)`
D. `(2f_(r)f_(a))/(f_(a) + f_(a))`

Answer

A

Explanation

`f_(a) = f(nu + nu_(o)/(nu))`
`:. nu_(o)/(nu )= f_(a)/(f) - 1` …(i)
`f_(r) = f ((nu - nu_(o))/(nu))`
`:. nu_(o)/(nu) = 1 - (f_(r))/(f)` …(ii)
From Eqs. (i) and (ii), we get
`f = (f_(a) + f_(r))/(2)`