The KE and PE of a particle executing SHM with amplitude A will be equal when its displacement is:

The KE and PE of a particle executing SHM with amplitude A will be equal when its displacement is:

A
$A \sqrt{2}$
B
$\frac{A}{2}$
C
$\frac{A}{\sqrt{2}}$
D
$A \sqrt{\frac{2}{3}}$

Solution:

$\frac{1}{2} m (A^{2} - y^{2}) ω^{2} = \frac{1}{2} {mω}^{2} y^{2} \Rightarrow A^{2} = 2 y^{2} \Rightarrow y = \frac{A}{\sqrt{2}}$

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