For a particle performing $S.H.M.$,the total energy is '$n$' times the kinetic energy,when the displacement of a particle from the mean position is $\frac{\sqrt{3}}{2} A$,where $A$ is the amplitude of $S.H.M.$ The value of '$n$' is

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 graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that

The displacement of a particle of mass $2 \,g$ executing $SHM$ is given by $y=5 \sin \left(4 t+\frac{\pi}{3}\right)$. Here,$y$ is in metres and $t$ is in seconds. The kinetic energy of the particle,when $t=\frac{T}{4}$ is (in $\,J$)

An object executes $SHM$ along the $x$-axis with an amplitude of $0.06 \,m$. At a certain distance '$x$' metres from the mean position,it has a kinetic energy of $10 \,J$ and a potential energy of $8 \,J$. The distance '$x$' will be: (in $\,m$)

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ is ......