$A$ particle undergoing simple harmonic motion has an amplitude of $10 \ cm$. When the particle is at a displacement of $6 \ cm$ from the centre,then the ratio of its kinetic energy to potential energy is

When a particle executing $SHM$ oscillates with a frequency $v$, then the kinetic energy of the particle

The displacement of a particle of mass $2 \text{ g}$ executing simple harmonic motion is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$,where $t$ is time in seconds. The maximum kinetic energy of the particle is (in $\text{ J}$)

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

$A$ linear harmonic oscillator of force constant $6 \times 10^5 \, N/m$ and amplitude $4 \, cm$ has a total energy of $600 \, J$. Select the correct statement.

$A$ particle is executing $S.H.M.$ with time period $T^{\prime}$. If the time period of its total mechanical energy is $T$,then $\frac{T^{\prime}}{T}$ is ........