The ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude,the distance being measured from its equilibrium position,is:

The kinetic energy of a particle,executing simple harmonic motion is $16 \ J$ when it is in the mean position. If the amplitude of motion is $25 \ cm$ and the mass of the particle is $5.12 \ kg$,the period of oscillation is

$A$ block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant $100 \text{ Nm}^{-1}$ whose other end is fixed to a rigid vertical wall. Initially, the block is at its equilibrium position. The block is pulled to a distance of $8 \text{ cm}$ and released. The kinetic energy of the block when it is at a distance of $3 \text{ cm}$ from the mean position is (in $\text{ J}$)

$A$ spring-mass system performs $S.H.M.$ If the mass is doubled keeping the amplitude same,then the total energy of $S.H.M.$ will become:

$A$ block is in simple harmonic motion $(S.H.M)$ at the end of a spring with position given by $x = 5 \cos \left(\omega t + \frac{\pi}{4}\right)$. If the total mechanical energy is $100 \ J$, then the potential energy at time $t = 0$ is: (in $J$)

Consider a simple harmonic motion $(SHM)$. Let $K$ and $U$ be kinetic energy and potential energy when the displacement in $SHM$ is one-half $\left(\frac{1}{2}\right)$ the amplitude. Which of the following statements is correct?