An object producing a pitch of $1200 H z$ is moving with a velocity of $50 m / s$ towards a stationary person. The velocity of sound is $350 m / s$ . The frequency of sound heard by the stationary person is :

1250 H z

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1050 H z

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700 H z

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1400 H z

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Solution

The correct option is B $1400 H z$
If $n_{a}$ be the apparent frequency, then

$n_{a} = n \times \frac{v_{s}}{(v_{s} - v_{0})}$

$= \frac{1200 \times 350}{(350 - 50)}$

$= \frac{1200 \times 350}{300} = 1400 H z$

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An object producing a pitch of 1200Hz is moving with a velocity of 50m/s towards a stationary person. The velocity of sound is 350m/s. The frequency of sound heard by the stationary person is :

The correct option is B 1400HzIf na be the apparent frequency, thenna=n×vs(vs−v0) =1200×350(350−50) =1200×350300=1400 Hz

An object producing a pitch of $1200 H z$ is moving with a velocity of $50 m / s$ towards a stationary person. The velocity of sound is $350 m / s$ . The frequency of sound heard by the stationary person is :

The correct option is B $1400 H z$
If $n_{a}$ be the apparent frequency, then

$n_{a} = n \times \frac{v_{s}}{(v_{s} - v_{0})}$

$= \frac{1200 \times 350}{(350 - 50)}$

$= \frac{1200 \times 350}{300} = 1400 H z$