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?
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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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