For a particle executing $S.H.M.$,the potential energy $V$ versus displacement $x$ graph is given. The value of the restoring force constant is ..... $N/m$.

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

$A$ body is executing a linear $S.H.M.$ Its potential energies at the displacements $x$ and $y$ are $E_1$ and $E_2$ respectively. Its potential energy at displacement $(x+y)$ will be

For a particle $P$ performing $S.H.M.$, when displacement is $x$, potential energy and restoring force acting on it are denoted by $E$ and $F$ respectively. The relation between $x, E$, and $F$ is:

$A$ 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

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like: