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 $ P E=\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}$
$ 4 A^{2}-4 x^{2} =x^{2} $
$ 4 A^{2}-5 x^{2}=0 $
$ 4 A^{2}= 5 x^{2}$
$x^{2}=\frac{4}{5} \,A^{2}$
$x=\frac{2}{\sqrt{5}} \,A$