$A$ body of mass $1\,kg$ is executing simple harmonic motion. Its displacement $y$ (in $cm$) at $t$ seconds is given by $y = 6\sin(100t + \pi/4)$. Its maximum kinetic energy is ..... $J$.

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 amplitude of a particle executing $S.H.M.$ is $3 \,cm$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is (in $\,cm$)

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

The variations of kinetic energy $K(x)$,potential energy $U(x)$,and total energy $E$ as a function of displacement $x$ of a particle in $SHM$ are as shown in the figure. The value of $|x_0|$ is

$A$ particle is vibrating simple harmonically with an amplitude $a$. The displacement of the particle when its energy is half kinetic and half potential is