The displacement of a particle executing $SHM$ is given by $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$. If $T$ is the time period and the mass of the particle is $2 \text{ g}$, the kinetic energy of the particle when $t = \frac{T}{4}$ is given by (in $\text{ J}$)

The $K.E.$ and $P.E.$ of a particle executing $SHM$ with amplitude $A$ will be equal when its displacement is

$A$ particle starts from mean position and performs $S.H.M.$ with period $T = 6 \text{ s}$. At what time is its kinetic energy $50\%$ of its total energy (in $\text{ s}$)? (Given: $\cos 45^\circ = 1/\sqrt{2}$)

The maximum potential energy of a block executing simple harmonic motion is $25 \ J$. $A$ is the amplitude of oscillation. At $x = A / 2$,the kinetic energy of the block is $...............$ (in $J$)

An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

The frequency at which kinetic energy changes into potential energy in a simple harmonic motion ($S$.$H$.$M$.) with frequency $f$ is: