JEE PHYSICS MCQ: WAVES

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Analyzing +4 | -1

A tunning fork of frequency of

512 \mathrm{~Hz}

when sounded with unknown tunning fork produces 5 beats/sec. If arms of the unknown fork are filed then it produces only 3 beats / sec. Find the frequency of unknown tunning fork.

View answer and explanation

Correct option: A

517 Hz

Option A (517 Hz) is correct because filing the arms of the unknown fork increases its frequency, and the beat frequency decreases, indicating that the unknown frequency was initially higher than 512 Hz.

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

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

-direction. In the fundamental mode, time taken by the wave to reach the closed end of the air filled pipe is

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}

T

is the reverberation time of an auditorium of volume

V