The displacement of a particle executing simple harmonic motion is given by $x=2 \cos (t)$, where $t$ is the time in seconds. Then, the time period of the particle is:

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

The displacement of a particle from its mean position (in metre) is given by
$y = 0.2 \sin (10 \pi t + 1.5 \pi) \cos (10 \pi t + 1.5 \pi)$
The motion of the particle is

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

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:

$A$ 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