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 $\,m$)

In simple harmonic motion,the total mechanical energy of a given system is $E$. If the mass of the oscillating particle $P$ is doubled,then the new energy of the system for the same amplitude is

$A$ particle starts oscillating simple harmonically from its equilibrium position with time period $T$. At time $t = \frac{T}{12}$,the ratio of its kinetic energy to potential energy is $\left[\sin \frac{\pi}{3} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin \frac{\pi}{6} = \cos \frac{\pi}{3} = \frac{1}{2}\right]$.

When a longitudinal wave propagates through a medium,the particles of the medium execute simple harmonic oscillations about their mean positions. These oscillations of a particle are characterised by an invariant

$A$ particle starts its oscillation from the equilibrium position with time period $T$. Find the ratio of kinetic energy to potential energy of the particle at time $t = \frac{T}{6}$.

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 $\text{ J}$)