The equation of motion of a particle executing simple harmonic motion is $4 \frac{d^2 y}{dt^2}+\pi^2 y=0$,where $y$ is in metres and $t$ is in seconds. The time period of oscillation of the particle is (in $s$)

$A$ particle is executing simple harmonic motion with time period $2 \ s$ and amplitude $1 \ cm$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5 \ s$,then $\frac{D}{d}$ is:

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:

$A$ 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:

$A$ particle is executing simple harmonic motion $(SHM)$ with a time period $T = 4 \ s$. At $t = 0$,the particle is at a position where its angular position is $45^\circ$ with the $x$-axis. Find the displacement equation of the particle along the $x$-axis.

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: