The ratio of kinetic energy at the mean position to potential energy at A 2 of a particle performing SHM is

Kinetic energy K = 1 2 m ω 2 (A 2 - y 2 ) At mean position y = 0 K = 1 2 m ω 2 ( A 2 ) Potential energy U = 1 2 m ω 2 y 2 ...(i) U a t y = A 2 U = 1 2 m A 2 4 ω 2 ...(ii) Dividing Eq. (i) by Eq. (ii), we get K E U = 4 1

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The ratio of kinetic energy at the mean position to potential energy at $\frac{A}{2}$ of a particle performing SHM is

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Solution:

Kinetic energy $K = \frac{1}{2} m ω^{2} {(A}^{2} - y^{2}$ )

At mean position $y = 0$

$K = \frac{1}{2} m ω^{2} (A^{2})$

Potential energy $U = \frac{1}{2} m ω^{2} y^{2}$ ...(i)

$U a t y = \frac{A}{2}$

$U = \frac{1}{2} \frac{m A^{2}}{4} ω^{2}$ ...(ii)

Dividing Eq. (i) by Eq. (ii), we get

$\frac{K E}{U} = \frac{4}{1}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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