What is the ratio of potential energy to the kinetic energy of a particle executing SHM when its displacement is half its amplitude?

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When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ' $a$ ' is(a) $a / 4$ (b) $a / 3$ (c) $a / 2$ (d) $2 a / 3$

When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______.

When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is 3 : 1.

Explanation:

x = A cos ωt

x = `"A"/2`

we get ωt = 60°

`"KE"/"PE" = (1/2 "m"ω^2"A"^2 sin^2 ω"t")/(1/2 "m"ω^2"A"^2 cos ω"t")`

= tan2 ωt

= tan2 60°

= 3

Concept: Energy in Simple Harmonic Motion

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