Question

A train approaching a hill at a speed of 40 km/hr sounds a whistle of frequency 580 hz when it is at a distance of 1 km from the hill. A wind with a speed of 40 km/hr is blowing in te direction of motion of train. What will be the frequency of the whistle as heard by an observer on the hill. ( velocity of sound in air = 1200 km/hr)

Answer 1

Answer:

n′=599.33Hz

Explanation:

According to Doppler's effect the apparent frequency when both source and observer move alond the same direction is

n’ = ((v+w)-v0 )/( (v+w)-vS ) x n

. Where v is velocity of sound in air, w is speed of wind, Vs is speed of train

 .

According to problem, velocity of observer is v0=0, so the equation becomes;

n’ = ((v+w)  )/( (v+w)-vS ) x n

Here v=1200kmh, w=40kmh, vS=40kmh and n=580Hz

Putting values in equation;

n’ = ((1200+40)  )/( (1200+40)-40 ) x 580

n′=599.33Hz