Question

A particle is vibrating in simple harmonic motion with amplitude of $4cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

• A. $1cm$
• B. $\sqrt{2}cm$
• C. $2cm$
• D. $2\sqrt{2}cm$

Answer 1

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

$Kineticenergy=\frac{1}{2}m{w}^2\left(A^2-x^2\right)$

Where x is the displacement from mean position

$\begin{array}{c}Potentialenergy=\frac{1}{2}m{w}^2{x}^2\\ \frac{1}{2}m{w}^2{x}^2=\frac{1}{2}m{w}^2\left(A^2-x^2\right)\\ x^2=A^2-x^2\\ x^2-x^2.A^2\\ 2x^2=4^2\\ 2x^2=16\\ x^2=8\\ x=2\sqrt{2}cm\end{array}$

Option D.