The kinetic energy of $SHM$ is $1/n$ times its potential energy. If the amplitude of the $SHM$ is $A$,what is the displacement of the particle?

$A$ body performs simple harmonic oscillations along the straight line $ABCDE$ with $C$ as the midpoint of $AE.$ Its kinetic energies at $B$ and $D$ are each one-fourth of its maximum value. If $AE = 2R,$ the distance between $B$ and $D$ is

$A$ vertical mass-spring system executes simple harmonic oscillations with a period of $2\, s$. $A$ quantity of this system which exhibits simple harmonic variation with a period of $1\, s$ is

$A$ particle is executing $S.H.M.$ of amplitude $A$. When the potential energy of the particle is half of its maximum value during the oscillation,its displacement from the equilibrium position is

The displacement of a particle executing $SHM$ is given by $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$. If $T$ is the time period and the mass of the particle is $2 \text{ g}$, the kinetic energy of the particle when $t = \frac{T}{4}$ is given by (in $\text{ J}$)

Write the maximum velocity of a $SHM$ oscillator in terms of mechanical energy $E$ and mass of oscillator $m$.