If the displacement equation of a particle is represented by $y = A\sin PT + B\cos PT$,the particle executes

$A$ potential is given by $V(x) = k(x+a)^2 / 2$ for $x < 0$ and $V(x) = k(x-a)^2 / 2$ for $x > 0$. The schematic variation of the oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is:

The displacement of a particle along the $x$ axis is given by $x = a \sin^2 \omega t$. The motion of the particle corresponds to

Match the following functions with their corresponding nature of motion, where $\omega$ is a constant:

List-$I$	List-$II$
$A$. $\sin^2 \omega t$	$I$. Periodic but not $SHM$ $(T = 2\pi/\omega)$
$B$. $\sin^3 \omega t$	$II$. Periodic but not $SHM$ $(T = \pi/\omega)$
$C$. $\sin \omega t + \cos \pi \omega t$	$III$. Non-periodic
$D$. $\cos \omega t + \cos 2\omega t$	$IV$. Periodic but not $SHM$ $(T = 2\pi/\omega)$

$A$ 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:

The periodic time of $SHM$ is also measured by $T = 2 \pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}$. Can we say that the time period depends on the displacement?