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  • A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be<div class='qna-opti

A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

Option: 1

1 : 1


Option: 2

1 : 3


Option: 3

1 : 4


Option: 4

2 : 1


Answers (1)

best_answer

Ratio of PE and KE = ?

When  \underset{\substack{\uparrow \\ \text { disp. }}}{\mathrm{x}}=\frac{\mathrm{A}}{2}

\begin{aligned} \mathrm{KE} & =\frac{1}{2} \mathrm{~m} \omega^2\left[\mathrm{~A}^2-\mathrm{x}^2\right] \\ & =\frac{1}{2} \mathrm{~m} \omega^2\left[\mathrm{~A}^2-\frac{\mathrm{A}^2}{4}\right] \end{aligned}

\mathrm{KE}==\frac{1}{2} \mathrm{~m} \omega^2\left[\frac{3 \mathrm{~A}^2}{4}\right]\, \, \, \, \, \, \, \, \, \, \, \, ....(1)

And \begin{aligned} \mathrm{PE} & =\frac{1}{2} m \omega^2 \mathrm{x}^2 \\ & =\frac{1}{2} \mathrm{~m} \omega^2 \frac{\mathrm{A}^2}{4} \end{aligned}

\begin{aligned} & \frac{\mathrm{PE}}{\mathrm{KE}}=\frac{\mathrm{m} \omega^2 \mathrm{~A}^2 \times 8}{8 \mathrm{~m} \omega^2 \mathrm{~A}^2 \times 3}=\frac{1}{3} \\ \Rightarrow & 1: 3 \end{aligned}

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Divya Prakash Singh

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