A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the appparent frequency heard by a scooter-driver approaching the policeman at a speed of 36.0 km/h ? Speed of sound in air = 340 m/s.

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Solution

u=Velocity of sound

$v_{m}$ =velocity of the mediam

$v_{0}$ =velocity of the observer

$v_{s}$ =velocity of the sources

$f = (\frac{\to u +_{m} -_{0}}{v + v_{m} - v_{s}}) F$

Using sign conventions in Doppler's effect

$v_{m} = 0, u = 340 m / s,$

$v_{s} = 0$ and $_{0} = - 10 m$

(36 km/h=10 m/s)

$f = (\frac{340 + 0 - 9 - 10}{340 + 0 - 0}) \times 2 k H z$

$= \frac{350 \times 2}{340} k H z$

=2.06 kHz.

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A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the appparent frequency heard by a scooter-driver approaching the policeman at a speed of 36.0 km/h ? Speed of sound in air = 340 m/s.

u=Velocity of sound vm =velocity of the mediam v0 =velocity of the observer vs =velocity of the sources f=(→u+→vm−→v0v+vm−vs)F Using sign conventions in Doppler's effect vm=0,u=340m/s, vs=0 and →v0=−10m (36 km/h=10 m/s) f=(340+0−9−10340+0−0)×2kHz =350×2340kHz =2.06 kHz.