Question

A particle executes SHM with an amplitude 4 cm. At what displacement from the mean position its energy is half kinetic and half potential?

Answer 1

The Displacement at which the KE = PE in  S.H.M = 71 % of Amplitude .
 Therefore the answer = 71/100 *4
                               = 2.84 cm
Hope This helps You!!

Answer 2

Explanation:

The particle executing the SHM that is Simple Harmonic Motion follows given below formulas,

$\text{Displacement}, \mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t})$

Also,

$\text { Velocity, } \mathrm{v}=-\mathrm{A} \omega \sin (\omega \mathrm{t})$

Thereby, we know that,

$\text {Maximum Kinetic Energy} =\frac{1}{2} m v_{\max }^{2}$

$\text {Maximum Kinetic Energy} =\frac{1}{2} m A^{2} \omega^{2}$

By the law of conservation of energy,

When kinetic energy is maximum, potential energy is zero thereby, when kinetic energy is half of its maximum, potential energy will be other half, implying that both will be equal at that instant.

Therefore,

Velocity of particle in SHM, at the position when kinetic energy is half of its maximum will be

$\frac{1}{2} m v^{2}=\frac{1}{2} \times \frac{1}{2} \times m \times A^{2} \times \omega^{2}$

$v=\frac{1}{\sqrt{2}} A \omega$

$v=A \omega \cos \omega t=\frac{1}{\sqrt{2}} A \omega$

$\omega t=\frac{\pi}{4}$

Thereby, $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}$

$x=\frac{A}{\sqrt{2}}$

Hence proved.