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

$A$ particle executes simple harmonic motion according to the equation $x(t) = A \sin^2(\alpha t)$. If the time period of the $SHM$ is $0.2 \ s$, then the value of $\alpha$ (in units of $rad/s$) is (in $pi$)

Which of the following functions of time represent $(a)$ periodic and $(b)$ non-periodic motion? Give the period for each case of periodic motion $[\omega$ is any positive constant].
$(i)$ $\sin \omega t + \cos \omega t$
$(ii)$ $\sin \omega t + \cos 2\omega t + \sin 4\omega t$
$(iii)$ $e^{-\omega t}$
$(iv)$ $\log(\omega t)$

$A$ particle is executing simple harmonic motion with an amplitude of $2 \,m$. The difference in the magnitudes of its maximum acceleration and maximum velocity is $4$. The time-period of its oscillation and its velocity when it is $1 \,m$ away from the mean position are respectively:

The displacement of a particle performing $S.H.M.$ is given by $Y = A \cos [\pi(t + \phi)]$. If at $t = 0$,the displacement is $y = 2 \text{ cm}$ and velocity is $v = 2\pi \text{ cm/s}$,the value of amplitude $A$ in $\text{cm}$ is:

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