A particle undergoes simple harmonic motion having an amplitude of 10 cm . At what distance from the mean position will its kinetic and potential energies be equal?

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Quantrex Academy · Accepted Answer

Correct answer: D. 5 2 cm

Let the distance from the mean position be

x

.

The kinetic energy of a particle in SHM is given by

K = \dfrac{1}{2} m \omega^2 (A^2 - x^2)

.

The potential energy is given by

U = \dfrac{1}{2} m \omega^2 x^2

.

Given that kinetic and potential energies are equal,

K = U

.

\dfrac{1}{2} m \omega^2 (A^2 - x^2) = \dfrac{1}{2} m \omega^2 x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \dfrac{A}{\sqrt{2}}

Substituting the amplitude

A = 10 \text{ cm}

:

x = \dfrac{10}{\sqrt{2}} = 5\sqrt{2} \text{ cm}

A particle undergoes simple harmonic motion having an amplitude of...

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