What is the ratio of the kinetic energy at the mean position to the potential energy at $y = A / 2$ for a particle performing $SHM$ (in $: 1$)?

$A$ 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 $J$)

The amplitude of a particle executing simple harmonic motion is $6 \ cm$. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes $4:5$ is (in $cm$)

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 $\,m$)

$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?

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 $\text{ J}$)