Question

In a SHM, potential energy of a particle at mean position is $E_1$ and kinetic energy is $E_2,$ then

• A. $E_1=E_2$
• B. $\text{total potential energy at}x=\frac{\sqrt{3}A}{2}\text{is}{E}_1+\frac{3}{4}{E}_2$
• C. $\text{total potential energy at}x=\frac{\sqrt{3}A}{2}\text{is}\frac{3}{4}{E}_2$
• D. $\text{None of these}$

Answer 1

The correct option is B $\text{total potential energy at}x=\frac{\sqrt{3}A}{2}\text{is}{E}_1+\frac{3}{4}{E}_2$

Potential energy of the particle executing SHM is given by,
$U-U_{mean}=\frac{1}{2}k{x}^2.........\left(1\right)$
From the data given in the question,
$U_{mean}=E_1$
Kinetic of the particle at the mean position is given by
$E_2=\frac{1}{2}k{A}^2$ {maximum kinetic energy}
So, at $x=\frac{\sqrt{3}}{2}A$
from $\left(1\right)$ we get,
$U-E_1=\frac{1}{2}k\times {\left(\frac{\sqrt{3}A}{2}\right)}^2$
$\Rightarrow U=E_1+\frac{1}{2}k{A}^2\times \frac{3}{4}\Rightarrow U=E_1+\frac{3}{4}{E}_2$
Thus, option (b) is the correct answer.