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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Solution

u=334 m/s

$v_{b} = \frac{4}{2} m / s, v_{0} = 0$

So, $v_{s} = v_{b} c o s θ$

$= 4 \times \sqrt{2} \times (\frac{1}{\sqrt{2}}) = 4 m / s$

So, the apparent frequency,

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

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

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u=334 m/s vb=42m/s,v0=0 So,vs=vb cos θ =4×√2×(1√2)=4m/s So, the apparent frequency, f=(u+0u−vb cos θ)f =(334334−4)×1650=1670Hz

u=334 m/s

$v_{b} = \frac{4}{2} m / s, v_{0} = 0$

So, $v_{s} = v_{b} c o s θ$

$= 4 \times \sqrt{2} \times (\frac{1}{\sqrt{2}}) = 4 m / s$

So, the apparent frequency,

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

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