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JEE PHYSICS MCQ: OSCILLATIONS

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A body of mass

10 \mathrm{~g}

is executing simple harmonic motion about a point with an amplitude

20 \mathrm{~cm}

. If its maximum velocity is

100 \mathrm{~cm}.\mathrm{s}^{-1}

, its velocity will be

50 \mathrm{~cm} .\mathrm{s}^{-1}

at a distance ____

\mathrm{cm}

from its mean position.

View answer and explanation

Correct option: A

10 \sqrt{3}

Option A (

10 \sqrt{3}

) is correct because it is calculated using the SHM velocity formula

v = \omega \sqrt{A^2 - y^2}

where

y

is the distance from the mean position.

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