Q. For any S.H.M. amplitude is $6\ cm.$ If instantaneous potential energy is half the total energy then distance of particle from its mean position is

A) $3\ cm$
B) $4.2\ cm$
C) $5.8\ cm$
D) $6\ cm$
Explanation

(b) If at any instant displacement is $y$ then it is given that
$U = \frac{1}{2} \times E \Rightarrow \frac{1}{2}m\omega^{2}y^{2} = \frac{1}{2} \times \left( \frac{1}{2}m\omega^{2}a^{2} \right)$
$\Rightarrow y = \frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 4.2\ cm$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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