Oscillations

JEE Main level, single correct, medium

80 of 194

Question 80 of 194

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

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Answer: (B) 1:31:3

The formula to calculate the potential energy of the particle executing SHM is given by
P=12mω2x2...(1)P=\dfrac{1}{2}m{\omega }^{2}{x}^{2}...\left(1\right)
The formula to calculate the kinetic energy of the same particle is given by
K=12mω2(A2−x2)...(2)K=\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right)...\left(2\right)
Divide equation (1) by equation (2) to obtain the required ratio.
PK=12mω2x212mω2(A2−x2)=x2A2−x2...(3)\begin{matrix}\dfrac{P}{K} & = & \dfrac{\dfrac{1}{2}m{\omega }^{2}{x}^{2}}{\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right)} \\ & = & \dfrac{{x}^{2}}{{A}^{2}-{x}^{2}}...\left(3\right)\end{matrix}
Substitute A2\dfrac{A}{2} for xx into equation (3) to calculate the required ratio.
PK=(A2)2A2−(A2)2=13\begin{matrix}\dfrac{P}{K} & = & \dfrac{{\left(\dfrac{A}{2}\right)}^{2}}{{A}^{2}-{\left(\dfrac{A}{2}\right)}^{2}} \\ & = & \dfrac{1}{3}\end{matrix}

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