A whistle giving out $450\,\,Hz$ approaches a stationary observer at a speed of $33\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$. The frequency heard by the observer in $Hz$ is [velocity of sound in air =$333\,\,{\rm{m}}{{\rm{s}}^{ - 1}}$

JEE PHYSICS MCQ: WAVES

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A whistle giving out

450\,\,Hz

approaches a stationary observer at a speed of

33\,\,\mathrm{m}{\mathrm{s}^{ - 1}}

. The frequency heard by the observer in

Hz

is [velocity of sound in air =

333\,\,\mathrm{m}{\mathrm{s}^{ - 1}}

View answer and explanation

Correct option: D

500

Option D (500) is correct because it is the frequency heard by the observer calculated using the Doppler effect formula when the source is moving towards the observer.

More Questions from WAVES

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The source of sound generating a frequency of

3\,\,kHz

reaches an observer with a speed of

0.5

times, the velocity of sound in air. The frequency heard by the observer is A source of sound emits waves with frequency

f\mathrm{ }Hz

and speed

V\mathrm{ }m/sec

. Two observers move away from this source in opposite directions each with a speed

0.2\,\,V

relative to the source. The ratio of frequencies heard by the two observers will be Assertion: The wave function of a pulse is given by

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

it propagates in

(– \mathrm{ve}) \,x

-direction.
Reason: The given wave function is of the form

y = f(kx +\omega t),

which represent a wave travelling in negative

X

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0.01\,\,s

. the fundamental frequency is A rope

5 \mathrm{~m}

long has a total mass of

245 \mathrm{~g}

. It is stretched with a constant tension of

1 \mathrm{~kg}

wt. If it is fixed at one end and shaken by hand at the other end, what frequency of shaking will make it break up into three vibrating segments? Take

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

. If

T

is the reverberation time of an auditorium of volume

V

then