$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. What is the ratio of potential energy to kinetic energy of the particle at time $t = \frac{T}{12}$? (Given: $\sin(\frac{\pi}{6}) = \frac{1}{2}$)

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation,the displacement of the particle from the equilibrium position in terms of its amplitude $a$ is

The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U(x)$ versus time $t$ plot of the particle is correctly shown in figure:

If $T$ is the period of $SHM$,then write the period of kinetic and potential energy.

$A$ particle oscillates along the $x$-axis according to the law,$x(t) = x_0 \sin^2\left(\frac{t}{2}\right)$,where $x_0 = 1 \text{ m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph.