The displacement of a particle is given by the relation $x = 4(\cos \pi t + \sin \pi t)$. The amplitude of the particle is

$A$ particle moves in $S.H.M.$ such that its acceleration is $a = -px$,where '$x$' is the displacement of the particle from the equilibrium position and '$p$' is a constant. The period of oscillation is

The displacement of a particle executing linear $S.H.M.$ is given by $x = 0.25 \sin(11t + 0.5) \ m$. The period of $S.H.M.$ is (take $\pi = \frac{22}{7}$):

In a linear simple harmonic motion $(SHM)$:
$(A)$ Restoring force is directly proportional to the displacement.
$(B)$ The acceleration and displacement are opposite in direction.
$(C)$ The velocity is maximum at the mean position.
$(D)$ The acceleration is minimum at extreme points.
Choose the correct answer from the options given below:

The function $\sin^2(\omega t)$ represents:

The angular frequency of motion whose equation is $4\frac{d^2y}{dt^2} + 9y = 0$ is ($y =$ displacement and $t =$ time).