Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator.

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

This is a short answer type question as classified in NCERT Exemplar

Let us assume that the required displacement be x

Potential energy of the simple harmonic oscillator =1/2 kx²

k= force constant=mw²

PE= ½ mw²x²

Maximum energy of oscillator

TE= ½ mw²A²

PE=1/2 TE

½ mw²x²= $\frac{1}{2} [\frac{1}{2} m w^{2} A^{2}]$

So x= $\sqrt[]{\frac{A^{2}}{2}}$ = $\pm \frac{A}{\sqrt[]{2}}$

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