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
B
2
C
8
D
4

Solution

## The correct option is C ZeroThe 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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