$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4 \, cm$ and a time period of $1.2 \, s$. The minimum time taken by the particle to move from $x = 2 \, cm$ to $x = +4 \, cm$ and back again is given by .... $s$.

$A$ particle executes a simple harmonic motion with a periodic time $8 \text{ s}$. At time $t = 0$, it is at a mean position. The ratio of the distance traveled by a particle in the $2^{nd}$ second and that in the $1^{st}$ second of its motion is $(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1)$.

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t)$?

The diagram shows two oscillations. What is the phase difference between the oscillations?

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is: