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 $\,m$)

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

$A$ particle undergoing simple harmonic motion has time-dependent displacement given by $x(t) = A \sin \left( \frac{\pi t}{90} \right)$. The ratio of kinetic energy to potential energy of the particle at $t = 210 \ s$ will be:

The kinetic energy,$K$ of a body performing simple harmonic motion varies with time $t$,is indicated in graph

The displacement of a particle in simple harmonic motion $(SHM)$ is given by $y = \sqrt{3 \pi} \sin \left(\frac{100}{\pi} t + \frac{\pi}{4}\right)$. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?