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

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) $$x={ \pm{A\over\sqrt2}}$$
(4)$$x={A\over2}$$

verified

The correct option is (3) $$x={ \pm{A\over\sqrt2}}$$

Explaination::

potential energy =$${1\over2}kx^2={1\over2}m\omega^2( x^2)$$

Kinetic energy =$${1\over2}m(\omega\sqrt{ A^2-x^2})^2$$

As ,

potential energy = kinetic energy

$${1\over2}m\omega^2( x^2)={1\over2}m(\omega\sqrt{A^2-x^2})^2$$

$${1\over2}m\omega^2( x^2)={1\over2}m\omega^2( A^2)-{1\over2}m\omega^2( x^2)$$

$$m\omega^2( x^2)={1\over2}m\omega^2A^2$$

$$x^2={1\over2}A^2$$

$$x=\pm{A\over\sqrt2}$$