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

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 ′ = n 0 ( v v − v s c o s θ ) n ′ = n 0 ( 300 300 − 30 c o s θ ) W h e n θ = π 2 . i.e., at N, n ′ = n 0 = 1000 H z , i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency n ′ = n 0 ( 300 300 − 30 c o s θ ) W h e n θ = 0 , i . e . , c o s θ = 1 , n m a x = n 0 v v − v s = ( 1000 H z ) ( 300 m / s ) ( 300 m / s − 30 m / s ) = 10 9 × 1000 H z = 1111 H z n m i n = n 0 v v + v s = 1000 × 300 330 = 909 H z

A stationary observer receives a sound from a sound of frequency v0 moving with

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Question -

A stationary observer receives a sound from a sound of frequency $v_{0}$ 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: Option A
:
A
This frequency–time curve corresponds to a source moving at an angle to a stationary observer.

A Stationary Observer Receives A Sound From A Sound Of Frequ...

In the region SN, the source is moving towards the observer, i.e., the apparent frequency

n^{'} = n_{0} (\frac{v}{v - v_{s} c o s θ})

n^{'} = n_{0} (\frac{300}{300 - 30 c o s θ})

W h e n θ = \frac{π}{2} .

i.e., at N,

n^{'} = n_{0} = 1000 H z,

i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency

n^{'} = n_{0} (\frac{300}{300 - 30 c o s θ})

W h e n θ = 0, i . e ., c o s θ = 1,

n_{m a x} = n_{0} \frac{v}{v - v_{s}} = \frac{(1000 H z) (300 m / s)}{(300 m / s - 30 m / s)}

= \frac{10}{9} \times 1000 H z = 1111 H z

n_{m i n} = n_{0} \frac{v}{v + v_{s}} = \frac{1000 \times 300}{330} = 909 H z

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