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A particle is vibrating in a simple harmonic motion with amplitude 3 cm. When the restoring force acting on the particle is half of the maximum possible restoring force, find the ratio of kinetic energy of the particle to its potential energy.

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$2 \sqrt{2}$

$\sqrt{3}$

answer is B.

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Detailed Solution

Restoring force acting on the particle $F = m ω^{2} x$ where x is the distance from equilibrium position. So $F_{\max} = m ω^{2} A$ and when $F = \frac{1}{2} F_{\max}, x = \frac{A}{2}$

$∴$ Kinetic energy $(K) = \frac{1}{2} m {[ω \sqrt{A^{2} - {(\frac{A}{2})}^{2}}]}^{2}$

$= \frac{3}{8} m ω^{2} A^{2}$

Potential energy (V) $= \frac{1}{2} m ω^{2} {(\frac{A}{2})}^{2} = \frac{1}{8} m ω^{2} A^{2}$

$∴ \frac{K}{U} = 3$

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