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 is vibrating in $S.H.M.$ 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 $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

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 $\,J$)

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

The total energy of a particle,executing simple harmonic motion is