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. $\frac{2}{4}$ A

Answer 1

Answer: (A)

Given, $\frac{KE}{PE}=\frac{1}{4}$
But we know that in $SHM$
$KE=\frac{1}{2}K\left(A^2-x^2\right)$
and $PE=\frac{1}{2}K{x}^2$
Hence, $\frac{\frac{1}{2}K\left(A^2-x^2\right)}{\frac{1}{2}K{x}^2}=\frac{1}{4}$
$\frac{A^2-x^2}{x^2}=\frac{1}{4}$
$4A^2-4x^2=x^2$
$4A^2-5x^2=0$
$4A^2=5x^2$
$x^2=\frac{4}{5}{A}^2$
$x=\frac{2}{\sqrt{5}}A$