If the function $\sin^2 \omega t$ (where $t$ is time in seconds) represents a periodic motion,then the period of the motion is:

For a simple harmonic motion given by $x = 0.01 \sin 100\pi (t + 0.05)$,what is the time period in $\text{sec}$?

$A$ particle of mass $0.1 \ kg$ is executing simple harmonic motion of amplitude $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. If the initial phase is $45^{\circ}$,the equation of its motion is (Assume $x(t)$ as the position of the particle at time $t$)

$A$ particle moves with simple harmonic motion in a straight line. In the first $\tau \ s$,after starting from rest,it travels a distance $a$,and in the next $\tau \ s$ it travels $2a$ in the same direction. Then:

$A$ point particle is acted upon by a restoring force $F = -k x^3$. The time period of oscillation is $T$ when the amplitude is $A$. The time period for an amplitude $2A$ will be

$A$ particle executing $SHM$ of amplitude $a$ has a displacement of $-\frac{a}{2}$ at $t = \frac{T}{4}$ and a positive velocity. Find the initial phase of the particle.