$A$ 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 ........ $cm$.

$A$ 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?

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 a particle executing $S.H.M.$,its potential energy is $8$ times its kinetic energy at a certain displacement $x$ from the mean position. If $A$ is the amplitude of $S.H.M.$,the value of $x$ is:

$A$ particle executes $SHM$ of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

$A$ block is executing $SHM.$ Let the time period of variation of velocity be $T_1$ and time period of variation of kinetic energy be $T_2.$ The relation between $T_1$ and $T_2$ is