$A$ particle is vibrating in a simple harmonic motion with an amplitude of $4\, cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

Write the coordinates of the points of intersection of the graph of kinetic energy $(K)$ and potential energy $(U)$ for a simple harmonic oscillator.

$A$ body of mass $1 \ kg$ is executing simple harmonic motion $(SHM)$. Its displacement $y$ (in $cm$) at time $t$ is given by $y = 6 \sin (100 t + \pi/4) \ cm$. Its maximum kinetic energy is (in $J$)

Which graph represents the difference between total energy and potential energy of a particle executing $SHM$ versus its distance from the mean position?

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 $: 1$)?

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .