The frequency of a tuning fork with amplitude $A=1 \mathrm{~cm}$ is $250 \mathrm{~Hz}$. The maximum velocity of any particle in air is equal to

$10 \mathrm{\pi} \mathrm{\textrm {ms } ^ { - 1 }}$

JEE PHYSICS MCQ: WAVES

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The frequency of a tuning fork with amplitude

A=1 \mathrm{~cm}

250 \mathrm{~Hz}

. The maximum velocity of any particle in air is equal to

View answer and explanation

Correct option: B

5 \mathrm{\pi} \mathrm{\textrm {ms } ^ { - 1 }}

Option B (

5 \pi \textrm{{ ms } ^ { - 1 }}

) is correct because it is calculated using the formula for maximum particle velocity in simple harmonic motion:

v_{max} = \omega A = 2\pi fA

Practice All Chapter MCQs →

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