Give the relation between periodic time and angular frequency.

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?

$A$ particle executing $SHM$ has a maximum speed of $0.5 \ m s^{-1}$ and a maximum acceleration of $1.0 \ m s^{-2}$. The angular frequency of oscillation is

$A$ simple harmonic motion having an amplitude $A$ and time period $T$ is represented by the equation: $y = 5 \sin \pi (t + 4) \ m$. Then the values of $A$ (in $m$) and $T$ (in $sec$) are:

If displacement $x$ and velocity $v$ are related as $4v^2 = 25 - x^2$ in a $SHM$. Then the time period of the given $SHM$ is (Consider $SI$ units).

Write the force law for $SHM$ and obtain the formula for the period of an $SHM$ particle.