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

Question

Dr. Abubacker · Accepted Answer

(d) Potential energy of particle
$U = \frac{1}{2}\ m\omega^{2}y^{2}$
Potential energy of maximum particle
$E = \frac{1}{2}\ m\omega^{2}A^{2}$
According to given position,
The potential energy $U = \frac{E}{2}$
or $\frac{1}{2}\ m\omega^{2}y^{2} = \frac{1}{2} 	imes \frac{1}{2}\ m\omega^{2}A^{2}$
$y^{2} = \frac{A^{2}}{2}$
$y = \ \frac{A}{\sqrt{2}}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?