Question

For what instant of time, does potential energy becomes equal to kinetic energy for a particle executing SHM with a time period T

• A. $T/4$
• B. $T/8$
• C. $T/2$
• D. $T$

Answer 1

The correct option is B $T/8$

The potential energy = $k{A}^{2}si{n}^2\omega t$ and
the kinetic energy = $k{A}^{2}co{s}^2\omega t$
If KE = PE, then it is possible only if $\omega t=\pi /4⟹t=T/8$
The correct option is (b)