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A small source of sound vibrating at frequency $$ 500\,Hz $$ is rotated in a circle of radius $$ 100/ \pi \, cm $$ at a constant angular speed of $$ 5.0 $$ revolutions per second. The speed of sound in air is $$ 330\, m/s $$.
For an observer who is at rest at a great distance from the centre of the circle but nearly in the same plane , the minimum $$ f_{min} $$ and the maximum $$ f_{max} $$ of the range of values of the apparent frequency heard by him will be 


A
fmin=455Hz,fmax=535Hz
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B
fmin=485Hz,fmax=515Hz
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C
fmin=485Hz,fmax=500Hz
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D
fmin=500Hz,fmax=515Hz
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Solution

The correct option is C $$ f_{min} = 485 \,Hz , f_{max} = 515\,Hz $$
Since the observer is in the same plane as the circle , at one instant , the source will move directly towards the observer when the apparent frequency will be maximum given by 
                      $$ f_{max} = \left  ( \dfrac{V}{V - u}  \right  ) 500 = \left  ( \dfrac{330}{330 - u}  \right  ) 500 $$ 
Now , $$ u = r \omega = \left ( \dfrac{100}{\pi} \right ) 5 \times 2 \pi \times 10^{-2} \, m/s $$ 
               $$ = 10\,m/s $$ 
 $$ \therefore $$            $$ f_{max} = \left ( \dfrac{330}{330 - 10} \right ) 500 = \dfrac{330}{320} \times 500 = 515 \,Hz $$ 
 Similarly , at another instant , the source will move directly away from the observer with velocity $$ u = 10 \, m/s $$ . At this instant , the apparent frequency will be minimum given by 
               $$ f_{min} = \left (  \dfrac{330}{330 + 10}\right ) 500 = \dfrac{33}{34} \times 500 = 485 \,Hz $$

Physics

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