Question

A particle executes $\text{SHM}$ on a line $8\text{cm}$ long. Its $\text{K.E.}$ and $\text{P.E.}$ will be equal when its distance from the mean position is :

• A. $4\text{cm}$
• B. $2\text{cm}$
• C. $2\sqrt{2}\text{cm}$
• D. $\sqrt{2}\text{cm}$

Answer 1

The correct option is C $2\sqrt{2}\text{cm}$

Let $A$ be the amplitude of the given $\text{SHM}$.
Length of the line in which $\text{SHM}$ executes will be twice of the amplitude of $\text{SHM}$.
So, $2A=8$
$\Rightarrow A=4\text{cm}$
From question,
$\text{K.E.}=\text{P.E.}$
$\Rightarrow \frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$
$\Rightarrow A^2-x^2=x^2$
$\Rightarrow 2x^2=A^2\Rightarrow x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\text{cm}$
Hence, option $\left(c\right)$ is the correct answer.