## In This post you will get answer of Q.

The kinetic energy of a particle executing simple harmonic motion is $ 1/6^{th} $ of its maximum value at a distance of $ 5 ,cm $ from its equilibrium position. The amplitude of its motion is

Solution:

## The kinetic energy of a particle executing simple harmonic motion at a distance $x$ from its equilibrium position is

$K=frac{1}{2}momega^{2} left(A^{2}-x^{2}right)$

where $m$ is the mass of the particle, $omega$ is the angular frequency and $A$ is the amplitude of oscillation

The kinetic energy is maximum at equilibrium position $(x=0)$ and its maximum value $(K_{0})$ is

$K_{0}=frac{1}{2} momega^{2},A^{2} dots(i)$

At $x=5,cm$,

$K=frac{1}{2}momega^{2} left(A^{2}-left(5,cmright)^{2}right)$

But $K=frac{1}{6}K_{0}$(given)

$therefore frac{1}{2}momega^{2}left(A^{2}-left(5,cmright)^{2}right)=frac{1}{6}left(frac{1}{2}momega^{2}A^{2}right)$

or $A^{2}-left(5,cmright)^{2}=frac{A^{2}}{6}$

or $A^{2}-frac{A^{2}}{6}=left(5,cmright)^{2}$

or $frac{5}{6}A^{2}=left(5 cmright)^{2}$

or $A=sqrt{frac{6}{5}}left(5,cmright)$

$=sqrt{30},cm$