Physics, asked by mechpari4397, 6 months ago

The frequency of whistle of an engine appears to be (4/5)th of initial frequency when it crosses a sationary observer.the velocity of the sound is 330m/s.then the speed of the engine is?

Answers

Answered by Adibajawed
3

Answer:

ANSWER

Given : Velocity of sound v

sound

=340m/s

Let the frequency of the whistle be n and velocity of the engine be v.

Let the apparent frequencies heard by the observer be n

and n

′′

before and after the source passes the observer.

Doppler effect when source (S

) approaches the stationary observer:

n

=n[

v

sound

−v

source

v

sound

] =n[

340−v

340

]

Doppler effect when source S moves away from the stationary observer:

n

′′

=n[

v

sound

+v

source

v

sound

] =n[

340+v

340

]

Now

n

′′

n

=

3

5

(Given)

340−v

340+v

=

3

5

Thus v=85m/s

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