Physics - SHM Question with Solution | TestHub

The potential energy of a particle executing S.H.M. is given by $U=\frac{1}{2}k{x}^2$ and the kinetic energy is given by $K=\frac{1}{2}k(A^2-x^2)$.

Given that $U=nK$, we have:

$\frac{1}{2}k{x}^2=n\left[\frac{1}{2}k(A^2-x^2)\right]$

$x^2=n(A^2-x^2)$

$x^2(1+n)=n{A}^2$

$\frac{n}{n+1}=\frac{x^2}{A^2}$

Substituting the given displacement $x=\frac{2\sqrt{2}}{3}A$:

$\frac{n}{n+1}={\left(\frac{2\sqrt{2}}{3}\right)}^2$

$\frac{n}{n+1}=\frac{8}{9}$

$9n=8n+8$

$n=8$

Answer: $8$