The mass of a particle is $1 \ kg$ and it is moving along the $x$-axis. The period of its oscillation is $\frac{\pi}{2} \ s$. Its potential energy at a displacement of $0.2 \ m$ is (in $J$)

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 graphs in the figure show that a quantity $y$ varies with displacement $d$ in a system undergoing simple harmonic motion. Which graph best represents the relationship obtained when $y$ is the total energy of the system?

The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in the figure. The oscillators have the same mass. The time period of oscillation is greatest for

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ mass $0.4 \,kg$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \,Hz$. At a certain displacement, it has kinetic energy $2 \,J$ and potential energy $1.2 \,J$. The amplitude of oscillation is (in $m$)