Question:
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?
Correct Option: , 3
Solution:
$\mathrm{KE}=\mathrm{PE}$
$\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}$
$A^{2}-x^{2}=x^{2}$
$2 x^{2}=A^{2}$
$x=\pm \frac{A}{\sqrt{2}}$
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