JEE PHYSICS MCQ: WAVES

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Application +4 | -1

A longitudinal progressive wave is given by the equation

y=5 \times 10^{-2} \sin \pi(400 {t}+{x}) {m}

. Find (i) amplitude (ii) frequency (iii) wave length and (iv) velocity of the wave. (v) velocity and acceleration of particle at

{x}=\frac{1}{6} {~m}

at

{t}=0.01

s (vi) maximum particle velocity and acceleration.

View answer and explanation

Correct option: A

8 \times 10^4 {~ms}^{-2}

Option A (

8 \times 10^4 {~ms}^{-2}

) is not the complete answer, but it contains the correct maximum acceleration. The question asks for multiple values, but this option only provides the maximum acceleration, and the other options are incorrect. The complete solution is in the steps below.

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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