The displacement $y(t) = A \sin (\omega t + \phi)$ of a pendulum for $\phi = \frac{2\pi}{3}$ is correctly represented by which of the following graphs?

$A$ 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 $(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1)$.

$A$ particle is executing $SHM$. The time taken for $\left(\frac{3}{8}\right)^{\text{th}}$ of an oscillation from extreme positions is $x$. Then,the time taken for the particle to complete $\left(\frac{5}{8}\right)^{\text{th}}$ of an oscillation from the mean position is

$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

Explain by drawing graphs of displacement $x(t) \to t$,velocity $v(t) \to t$ and acceleration $a(t) \to t$ of $SHM$ for initial phase zero.

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is: