  Question

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  B
fmin=485Hz,fmax=515Hz  C
fmin=485Hz,fmax=500Hz  D
fmin=500Hz,fmax=515Hz  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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