  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 $$330\, m/s$$.If the observer moves towards the source with a constant speed of $$20\,m/s$$ , along the radial line to the centre , the fractional change in the apparent frequency over the frequency that the source will have if considered at rest at the centre will be

A
6 %  B
3 %  C
2 %  D
9 %  Solution

The correct option is A $$6$$ %The frequency of the source considered stationary at the centre is $$n_0 = 500\,Hz$$  when the observer moves towards the centre with constant speed  $$u$$ , the apparent frequency is               $$n_a = \left ( \dfrac{V + u}{V} \right ) 500 = \dfrac{330 + 20}{330} \times 500 = \dfrac{35}{33} \times 500$$  Change in frequency is given by            $$n_a - n_0 = \left ( \dfrac{35}{33} \times 500 - 500 \right ) = \dfrac{2 \times 500}{33} \,hz$$  Fractional change of frequency is          $$\left (\dfrac{n_a - n_0}{n_0} \right ) = \left (\dfrac{2 \times 500}{33} \right ) \times \dfrac{1}{500}$$                                     $$= \dfrac{2}{33} = 0.06$$ Hence , percentage change is $$6$$ % Physics

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