For a particle executing $S.H.M.$,its potential energy is $8$ times its kinetic energy at a certain displacement $x$ from the mean position. If $A$ is the amplitude of $S.H.M.$,the value of $x$ is:

The potential energy of a simple harmonic oscillator at the mean position is $2\,J$. If its mean kinetic energy $(K.E.)$ is $4\,J$,its total energy will be .... $J$

$A$ particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ from the mean position,its kinetic energy is increased by an amount $\frac{1}{2}m\omega^2A^2$ due to an impulsive force. What is its new amplitude?

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 $\text{ J}$)

The displacement of a particle in simple harmonic motion $(SHM)$ is given by $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}\right)$. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

The displacement-time graph of a particle executing $SHM$ is as shown in the figure. The corresponding graph between potential energy $(PE)$ and time is: