$A$ 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 ........

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 $\,m$)

The position of a particle executing simple harmonic motion is given by $x(t) = 2 \cos \left(\frac{\pi}{15} t - \frac{\pi}{2}\right)$, where $x$ is in centimetre and $t$ is in seconds. The time period of the kinetic energy of the particle in seconds is

The kinetic energy of a particle executing simple harmonic motion in a straight line is $pv^2$ and the potential energy is $qx^2$,where $v$ is the speed at a distance $x$ from the mean position. Its time period is given by the expression:

$A$ body of mass $m$ is executing $SHM$ with amplitude $a$. When its displacement $x = 1$ unit,the force is $b$. What will be its maximum kinetic energy?

The variations of potential energy $(U)$ with position $x$ for three simple harmonic oscillators $A, B$ and $C$ are shown in the figure. The oscillators have the same mass. The time period of oscillation is greatest for