The amplitude of a particle executing $S.H.M.$ is $3 \,cm$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is (in $\,cm$)

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

$A$ particle executing simple harmonic motion has a kinetic energy $K = K_0 \cos^2(\omega t)$. The maximum values of the potential energy and the total energy are respectively:

In a simple harmonic oscillation,what fraction of total mechanical energy is in the form of kinetic energy,when the particle is midway between mean and extreme position?

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement (amplitude = $a$) is