Question

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude=a) is :

• A. $\frac{a}{2}$
• B. $a\sqrt{2}$
• C. $\frac{a}{\sqrt{2}}$
• D. $\frac{a\sqrt{2}}{3}$

Answer 1

The correct option is C $\frac{a}{\sqrt{2}}$

Suppose at displacement $y$ from mean position.

K.E $=\frac{1}{2}m(a^2-y^2){\omega }^2$

P.E $=\frac{1}{2}m{\omega }^2{y}^2$

potential energy = kinetic energy
$\Rightarrow \frac{1}{2}m(a^2-y^2){\omega }^2=\frac{1}{2}m{\omega }^2{y}^2$

$\Rightarrow a^2=2y^2\Rightarrow y=\frac{a}{\sqrt{2}}$