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Question
A source S of acoustic wave of the frequency v0=1700Hz and a receiver R are located at the same point. At the instant t=0, the source starts from rest to move away from the receiver with a constant acceleration ω. The velocity of sound in air is v=340m/s.
If ω=10m/s2, the apparent frequency that will be recorded by the stationary receiver at t=10s will be
A source S of acoustic wave of the frequency v0=1700Hz and a receiver R are located at the same point. At the instant t=0, the source starts from rest to move away from the receiver with a constant acceleration ω. The velocity of sound in air is v=340m/s.
If ω=10m/s2, the apparent frequency that will be recorded by the stationary receiver at t=10s will be
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Solution
The correct option is B 1314Hz
At t= 10s the velocity of the source is= 10×10=100m/sUsing the concept of Doppler effect
f′=f(v−vov−vs) where o is the observer and s is the source⇒f′=1700(340340−(−100))=1314Hz
At t= 10s the velocity of the source is= 10×10=100m/s
Using the concept of Doppler effect
f′=f(v−vov−vs) where o is the observer and s is the source
⇒f′=1700(340340−(−100))=1314Hz
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