Answer to Question #164374 in Optics for Gasty

Question

 An observer travelling with a constant velocity of 25ms-1 passes close to a stationary source and notices that there is a change of frequency of 60Hz as he passes the source. What is the

           frequency of the source? 

Answer 1

This is the application of Doppler's Effect.

According to Doppler's condition :

frequency increase as the distance is decreasing between the observer and source (or observer is approaching) due to their relative velocities.

or

frequency decreases as the distance is increasing between the observer and source (or observer is going away)due to their relative velocities.

This is condition mathematically given as follows:

$\boxed{f'=f({v\pm v_o \over v\pm v_s})}$

here, $v=$ velocity of sound in air (340 m/s)

$v_0=$ velocity of observer

$v_s=$ velocity of source

$f=$ frequency of source

$f'=$ observed frequency

in first condition given observer is approaching towards the source so frequency will increase:

$\therefore\boxed{f_1'=f({v+25 \over v})}$ .....................(1)

and in second condition observer is going away from the source so frequency will decrease

$\therefore\boxed{f_2'=f({v-25 \over v})} -$ .................(2)

after putting value of $f_2'=f_1'-60$ (given) and velocity of sound=340,we get;

$f_1'-f_2'=60\\ =f({365\over340})-f({315\over340})$

$60=f({50\over340})$

$\boxed{f=408H_Z}$ .....frequency of source (answer)