A particle executing simple harmonic motion along x-axis, with amplitude A, about origin. If ratio of kinetic energy and total energy at x = A/3 is

The correct option is 8/9.

Let us find the ratio of the kinetic energy (K.E.) to the total energy (T.E.) of a particle executing simple harmonic motion (SHM) at a displacement of $x=\frac{A}{3}$, where $A$ is the amplitude.

For a particle of mass $m$ executing SHM with angular frequency $\omega$ and amplitude $A$, the total energy ($E_{\text{total}}$) is constant at all positions and is given by the formula:
$E_{\text{total}}=\frac{1}{2}m{\omega }^2{A}^2$

The potential energy (P.E.) of the particle at a displacement $x$ from the mean position is given by:
$E_{\text{potential}}=\frac{1}{2}m{\omega }^2{x}^2$

The kinetic energy (K.E.) of the particle at any displacement $x$ is the difference between the total energy and the potential energy:
$E_{\text{kinetic}}=E_{\text{total}}-E_{\text{potential}}$
Substituting the expressions for total energy and potential energy:
$E_{\text{kinetic}}=\frac{1}{2}m{\omega }^2{A}^2-\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$

We are required to find the ratio of kinetic energy to total energy at $x=\frac{A}{3}$:
$\text{Ratio}=\frac{E_{\text{kinetic}}}{E_{\text{total}}}=\frac{\frac{1}{2}m{\omega }^2(A^2-x^2)}{\frac{1}{2}m{\omega }^2{A}^2}$
Simplifying the ratio, we get:
$\text{Ratio}=\frac{A^2-x^2}{A^2}=1-\frac{x^2}{A^2}$

Now, substitute $x=\frac{A}{3}$ into this relation:
$\text{Ratio}=1-\frac{{(A/3)}^2}{A^2}$
$\text{Ratio}=1-\frac{A^2/9}{A^2}$
$\text{Ratio}=1-\frac{1}{9}=\frac{8}{9}$

Thus, the ratio of kinetic energy to total energy at the given position is $\frac{8}{9}$.