$A$ particle starts executing simple harmonic motion $(SHM)$ of amplitude $a$ and total energy $E$. At any instant,its kinetic energy is $\frac{3E}{4}$. Then its displacement $y$ is given by:

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)

$A$ 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?

$A$ particle executes simple harmonic motion with a frequency $f$. The frequency with which its kinetic energy oscillates 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 .... $J$.

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ is ......