JEE PHYSICS MCQ: WAVES

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Application +4 | -1

A transverse sinusoidal wave of amplitude

A

, wavelength

\lambda

and frequency

n

is travelling on a stretched string. The maximum speed of particle is

(1 / 10)

th the speed of propagation of the wave. If

A=10^{-3} \mathrm{~m}

and

v=10 \mathrm{~ms}^{-1}

, then

\lambda

and

n

are given by

View answer and explanation

Correct option: A

\lambda=2 \pi \times 10^{-2} \mathrm{~m}

Option A (

\lambda=2 \pi \times 10^{-2} \mathrm{~m}

) is correct because it is the wavelength calculated using the given amplitude, frequency, and wave speed.

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3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

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

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

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

T

V