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The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = $a$) is
JEE MAIN 2021, Medium
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(c) Suppose at displacement $y$ from mean position potential energy = kinetic energy 

==> $\frac{1}{2}m({a^2} - {y^2}){\omega ^2} = \frac{1}{2}m{\omega ^{\rm{2}}}{y^2}$ 

==> ${a^2} = 2{y^2}$ 

==> $y = \frac{a}{{\sqrt 2 }}$

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