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 $\text{ J}$)

Starting from the origin,a particle oscillates simple harmonically with a time period of $2 \ s$. After what time will its kinetic energy be $75 \%$ of the total energy?

$A$ particle starts oscillating simple harmonically from its mean position with time period $T$. At time $t = T/6$,the ratio of the potential energy to kinetic energy of the particle is $\left[\sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2\right]$

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:

$A$ particle is vibrating in a simple harmonic motion with an amplitude of $4\, cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

$A$ particle of mass $m$ performs $SHM$ along a straight line with frequency $f$ and amplitude $A.$