Question

A particle is vibrating in S H M with an amplitude of 4 cm. At what displacement from the equilibrium position does it have half potential and half kinetic?

• A. 1 cm
• B. $\sqrt{2}cm$
• C. 2 cm
• D. 2$\sqrt{2}cm$

Answer 1

The correct option is D 2$\sqrt{2}cm$

In SHM the KE and PE of particle is given by the following relations:

$KE=\frac{1}{2}m{\omega }^2(a^2-x^2)$

$PE=\frac{1}{2}m{\omega }^2{x}^2$

So, the total energy,

$TE=KE+PE$

$TE=\frac{1}{2}m{\omega }^2{a}^2$

As, PE = KE (both half). From above equations:

$a^2-x^2=x^2$

$2x^2=a^2$

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

$x=\frac{4}{\sqrt{2}}$

$x=2\sqrt{2}cm$