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:

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

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 $\text{ J}$)

Consider a simple harmonic motion $(SHM)$. Let $K$ and $U$ be kinetic energy and potential energy when the displacement in $SHM$ is one-half $\left(\frac{1}{2}\right)$ the amplitude. Which of the following statements is correct?

The amplitude of a particle executing $SHM$ is $3 \text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than its potential energy is (in $\text{cm}$):

The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is