Question
A stationary observer receives a sound from a sound of frequency v0 moving with a constant velocity vs=30 m/s. The apparent frequency varies with time as shown in figure. Velocity of sound v = 300 m/s. Then which of the following is incorrect?
Answer: Option A
:
A
This frequency–time curve corresponds to a source moving at an angle to a stationary observer.
In the region SN, the source is moving towards the observer, i.e., the apparent frequency
n′=n0(vv−vscosθ)
n′=n0(300300−30cosθ)
Whenθ=π2. i.e., at N,
n′=n0=1000Hz, i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency
n′=n0(300300−30cosθ)
Whenθ=0,i.e.,cosθ=1,
nmax=n0vv−vs=(1000Hz)(300m/s)(300m/s−30m/s)
=109×1000Hz=1111Hz
nmin=n0vv+vs=1000×300330=909Hz
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:
A
This frequency–time curve corresponds to a source moving at an angle to a stationary observer.
In the region SN, the source is moving towards the observer, i.e., the apparent frequency
n′=n0(vv−vscosθ)
n′=n0(300300−30cosθ)
Whenθ=π2. i.e., at N,
n′=n0=1000Hz, i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency
n′=n0(300300−30cosθ)
Whenθ=0,i.e.,cosθ=1,
nmax=n0vv−vs=(1000Hz)(300m/s)(300m/s−30m/s)
=109×1000Hz=1111Hz
nmin=n0vv+vs=1000×300330=909Hz
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