A boy riding on his bike is going towards east at a speed of $4 \sqrt{2} m s^{- 1}$ . At a certain point he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of 334 $m s^{- 1}$ . A second boy stands on the ground 45 $^{\circ} C$ south of east from him. Find the frequency of the pulse as received by the second boy.

1670

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1640

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1650

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None of these

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Solution

The correct option is A 1670

$u = 334 m / s, v_{b} = 4 \sqrt{2} m / s, v_{o} = 0$

So, $v_{s} = v_{b} c o s θ = 4 \sqrt{2} \times (\frac{1}{\sqrt{2}}) = 4 m / s$ .

So,the apparent frequency $f^{'} = (\frac{u + 0}{u - v_{b} c o s θ}) f$

= $(\frac{334}{334 - 4}) 1650 = 1670 H z$ .

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Similar questions

A boy riding on his bike is going towards east at a speed of $4 \sqrt{2} m s^{- 1}$ . At a certain point he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of $334 m s^{- 1}$ . A second boy stands on the ground $45^{\circ}$ south of east from him. Find the frequency of the pulse as received by the second boy.

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The correct option is A 1670 u=334m/s,vb=4√2m/s,vo=0 So,vs=vb cos θ=4√2×(1√2)=4m/s. So,the apparent frequency f′=(u+0u−vbcosθ)f =(334334−4)1650=1670Hz.

The correct option is A 1670

$u = 334 m / s, v_{b} = 4 \sqrt{2} m / s, v_{o} = 0$

So, $v_{s} = v_{b} c o s θ = 4 \sqrt{2} \times (\frac{1}{\sqrt{2}}) = 4 m / s$ .

So,the apparent frequency $f^{'} = (\frac{u + 0}{u - v_{b} c o s θ}) f$

= $(\frac{334}{334 - 4}) 1650 = 1670 H z$ .