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 :

\[\left( \frac{1}{2} \right)m \omega^2 \left( A^2 - y^2 \right) = \left( \frac{1}{2} \right)m \omega^2 y^2\]

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.}