Question

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

Answer 1

It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm

To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
$\omega$ be the angular speed of the particle, and
A be the amplitude of the simple harmonic motion.

Equating the mathematical expressions for K.E. and P.E. of the particle, we get:
$\left(\frac{1}{2}\right)m{\mathrm{\omega }}^2\left(A^2-y^2\right)=\left(\frac{1}{2}\right)m{\mathrm{\omega }}^2{y}^2$
A² − y² = y²
2y² = A²

$\Rightarrow y=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$
The kinetic energy and potential energy of the particle are equal at a distance of $5\sqrt{2}$ cm from the mean position.