Question

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

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

Answer 1

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

Let equation of SHM is $x=Asin\left(\omega t\right)$, WHere $A$ is amplitude and $\omega$ is angular velocity.

$\Rightarrow v=A\omega cos\left(\omega t\right)$

Kinetic Energy(KE)$=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2\left(\omega t\right)=\frac{1}{2}m{A}^2{\omega }^2(1-si{n}^2(\omega t\left)\right)$

$\Rightarrow \text{Kinetic Energy}=\frac{1}{2}m{\omega }^2(A^2-x^2)$,

Total Energy = Kinetic Energy + Potential Energy

Total Energy at mean position= Kinetic Energy ($\because$ PE=0)

$\Rightarrow \text{Total Energy}=\frac{1}{2}m{A}^2{\omega }^2$

Let at position $x$ Kinetic Energy and potential Energy will be equal.

KE$=\frac{1}{2}m{\omega }^2(A^2-x^2)$

Potential Energy= Total Energy - Kinetic Energy$=\frac{1}{2}m{A}^2{\omega }^2-\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$

$\Rightarrow KE=PE$ $\Rightarrow \frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$ $\Rightarrow x=\frac{A}{\sqrt{2}}$

Given $A=4cm$ $\Rightarrow x=2\sqrt{2}cm$