$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.

If the frequency of a simple harmonic motion is $f$,what is the frequency of its kinetic energy?

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x+y)$ is

$A$ particle starts oscillating simple harmonically from its equilibrium position. The ratio of kinetic energy and potential energy of the particle at time $t = T/12$ is: ($T =$ time period)

$A$ vertical mass-spring system executes simple harmonic oscillations with a period of $2\, s$. $A$ quantity of this system which exhibits simple harmonic variation with a period of $1\, s$ is

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)