$A$ 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

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)

$A$ body is executing simple harmonic motion with frequency $n$,the frequency of its potential energy is:

Write the maximum velocity of a $SHM$ oscillator in terms of mechanical energy $E$ and mass of oscillator $m$.

$A$ body is performing $SHO$ with a total energy of $100\,J$. In the table below, column-$I$ shows the kinetic energy $(K)$ at a specific time, and column-$II$ shows the potential energy $(U)$ at that same time. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $K = 10\,J$	$(i)$ $U = 40\,J$
$(b)$ $K = 60\,J$	$(ii)$ $U = 90\,J$
	$(iii)$ $U = 50\,J$