$A$ 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 $s$)

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

The total energy of a body executing simple harmonic motion is $E$. When the displacement is half of the amplitude,the kinetic energy 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 $\text{ J}$)

$A$ body is performing simple harmonic motion with an amplitude of $10 \, cm$. The velocity of the body is tripled by an air jet when it is at $5 \, cm$ from its mean position. The new amplitude of vibration is $\sqrt{x} \, cm$. The value of $x$ is . . . . . . .

Find the displacement of a simple harmonic oscillator at which its $PE$ is half of the maximum energy of the oscillator.