$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

$A$ particle oscillates in a straight line simple harmonically with a period of $8 \ s$ and an amplitude of $4 \sqrt{2} \ m$. The particle starts from the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ second of its motion to that in the $2^{\text{nd}}$ second is: $(\sin 45^{\circ} = 1 / \sqrt{2}, \sin \frac{\pi}{2} = 1)$

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

Consider a pair of identical pendulums,which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical,the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums?

$A$ particle starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is