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Question

A boy is walking away from a wall towards an observer at a speed of 1 metre/sec and blows a whistle whose frequency is 690 Hz. The number of beats heard by the observer per second is (Velocity of sound in air = 340 metres/sec          [MP PMT 1995]


A
Zero
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B
2
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C
8
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D
4
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Solution

The correct option is C Zero
The apparent frequency due to doppler effect is given by, 
$$\nu^{\prime}=\nu\left(\dfrac{v\pm v_o}{v\pm v_s}\right)$$
In one case the source is moving $$away$$ from the observer. The sign used here will be $$+$$. We get,
$$\nu^{\prime}=\dfrac{\nu v}{v-v_s}$$ ($$\because$$ the observer is at rest)
Using the given values, 
$$\nu^{\prime}=\dfrac{680\times 340}{340+1}$$
In the other case the source is moving $$towards$$ the observer. The sign used will be $$-$$.
$$\nu^{\prime \prime}=\dfrac{680\times 340}{340-1}$$
The beat frequency, $$f_b=\nu^{\prime \prime}-\nu^{\prime}$$
or, $$f_b\approx 0$$

Physics

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