Question

For what value of displacement the kinetic energy and potential of a simple harmonic oscillation become equal ?

(1) x = 0 

(2) $x=\pm A$ 

(3) $x=\pm \frac{A}{\sqrt{2}}$ 

(4)$x=\frac{A}{2}$

Answer 1

The correct option is (3) $x=\pm \frac{A}{\sqrt{2}}$ 

Explaination::

potential energy =$\frac{1}{2}k{x}^2=\frac{1}{2}m{\omega }^2(x^2)$ 

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

As , 

potential energy = kinetic energy 

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

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

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

$x^2=\frac{1}{2}{A}^2$

$x=\pm \frac{A}{\sqrt{2}}$