Question

A particle is executing simple harmonic motion with amplitude $A$. When the ratio of its kinetic energy to the potential energy is $\frac{1}{4}$, its displacement from its mean position is

• A. $\frac{2}{\sqrt{5}}A$
• B. $\frac{\sqrt{3}}{2}A$
• C. $\frac{3}{4}A$
• D. $\frac{1}{4}A$
• E. $\frac{2}{5}A$

Answer 1

The correct option is A $\frac{2}{\sqrt{5}}A$

$K.E=\frac{1}{2}k(A^2-x^2)$
$P.E=\frac{1}{2}k{x}^2$
Given : $\frac{K.E}{P.E}=\frac{1}{4}$
$\frac{(A^2-x^2)}{x^2}=\frac{1}{4}$
Solving above we get,
$x=2A/\sqrt{5}$