A stationary observer receives a sound from a sound of freqeuency `v_0` moving with a constant velocity `v_S=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?

Options

A. The minimum value of apparent frequency is 889 Hz
B. The natural frequency of source is 1000 Hz
C. The frequency-time curve corresponds to a source moving at an angle to the stationary observer
D. The maximum value of apparent frequency is 1111 Hz

Answer

A

Explanation

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.
`f'=f_(0)((V)/(V-V_(S)costheta)), f'=f_(0)((300)/(300-30costheta))`
When `theta=(pi)/(2), ` i.e., at N,
`f'=f_(0)=1000Hz` i.e., natural frequency of source
In the region `NS'` the source is moving away from the observer,
i.e., apparent frequency
`f=f_(0)((300)/(300+30costheta))`
When `theta=(pi)/(2)`, i.e., At N,
`f'=f_(0)=1000Hz`, i.e., natural frequency of source.
In the region `NS'` the source is moving away from the observer, i.e., apparent frequency
`f'=f_(0)((300)/(300+30costheta))`
When `theta=0` i.e., `costheta=1`
`f_(max)=f_(0)(v)/(v-v_(S))=((100Hzxx300m//s))/(300m//s-30m//s))=(10)/(9)xx1000Hz=1111Hz`
`f_("min")=f_(0)(v)/(v+v_(S))=(1000xx300)/(330)=909Hz`