The acceleration $A$ and time $t$ of a body in $S.H.M.$ is given by the curve shown below. Then the corresponding graph between kinetic energy $(K.E.)$ and time $t$ is correctly represented by:

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 displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .

The ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude,the distance being measured from its equilibrium position,is:

$A$ 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:

The potential energy of a particle with displacement $X$ is $U(X)$. The motion is simple harmonic,when ($K$ is a positive constant)