Question

A particle of amplitude A is executing simple harmonic motion. When the potential energy of particle is half of its maximum potential energy, then displacement from its equilibrium position is



A/4

A/3

A/2

A/√2

​

Answer 1

▶ Solution :

Given :

▪ A particle is performing SHM.

▪ Amplitude of particle is A.

To Find :

Displacement of the particle from the mean position when the potential energy of particle is half of its maximum potential energy.

Formula :

Formula of PE of a particle performing SHM is given by

$\boxed{\bf{\pink{PE=\dfrac{1}{2}kX^2}}}$

• X denotes displacement from mean position
• k = mω$^2$

Calculation :

Maximum potential energy of particle in SHM is given by

$\implies\bf\red{(PE)_{max}=\dfrac{1}{2}kA^2}$

ATQ,

$\implies\sf\:PE=\dfrac{(PE)_{max}}{2}\\ \\ \implies\sf\:\dfrac{1}{2}kX^2=\dfrac{\frac{1}{2}kA^2}{2}\\ \\ \implies\sf\:X^2=\dfrac{A^2}{2}\\ \\ \implies\boxed{\bf{\orange{X=\dfrac{A}{\sqrt{2}}}}}$