$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

$A$ 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

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

The kinetic energy of a particle,executing simple harmonic motion is $16 \ J$ when it is in the mean position. If the amplitude of motion is $25 \ cm$ and the mass of the particle is $5.12 \ kg$,the period of oscillation is