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) \(x={ \pm{A\over\sqrt2}}\) 

(4)\(x={A\over2}\)

Answer 1

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 

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

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

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

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

\( x=\pm{A\over\sqrt2}\)