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 $^{\circ} C$ south of east from him. Find the frequency of the pulse as received by the second boy.

Q. A boy riding on a bicycle going at 12 km h⁻¹ towards a vertical wall whistles at his dog on the ground. If the frequency of the whistle is 1600 Hz and the speed of sound in air is 330 m s⁻¹, find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy.

Q. A man is travelling towards east at 20ms

^{- 1}

and sound waves are moving towards east at 340ms

^{- 1}

. He finds that

500

waves cross him in

2

seconds, the wavelength of the sound wave received by the man is :

Q. A boy is walking away from a wall at a speed of 1m/s in a direction at right angle to the wall. As he walks, he blows a whistle steadily. An observer towards whom the boy is walking hears 4 beats per second. If the speed of sound is 340m/s, what is the frequency of the whistle?

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

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