The amplitude of a particle executing $SHM$ is $3\,cm$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: $.............cm$.

$A$ simple pendulum of mass $m$ executes $S.H.M.$ with total energy $E$. If at an instant it is at one of the extreme positions,then its linear momentum after a phase shift of $\frac{\pi}{3} \, rad$ will be

$A$ 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:

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)

Starting from the origin,a body oscillates simple harmonically with a period of $2 \ s$. After what time will its kinetic energy be $75\%$ of the total energy?

The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)