Question

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic

[NCERT 1984; MNR 1995;

RPMT 1995; DCE 2000; UPSEAT 2000]

• A. 1 cm
• B.
• C. 3 cm
• D.

Answer 1

The correct option is D

Let x be the point where K.E = P.E

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

$\Rightarrow 2x^2=a^2\Rightarrow x=\frac{a}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$ cm