$A$ point particle of mass $200 \text{ g}$ is executing $S.H.M.$ of amplitude $0.2 \text{ m}$. When the particle passes through the mean position,its kinetic energy is $16 \times 10^{-3} \text{ J}$. The equation of motion of this particle is (Initial phase of oscillation $= 0^{\circ}$)

The equation of motion of a particle is $\frac{d^2y}{dt^2} + Ky = 0$,where $K$ is a positive constant. The time period of the motion is given by

Write the equation representing the relation between time period and frequency.

The displacement of a particle executing $S.H.M.$ is given by $x = 0.01 \sin 100 \pi(t + 0.05)$. The time period is ........ $s$.

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

$A$ ball is released from height $h$ which makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is ($g$ = acceleration due to gravity).