$A$ 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:

Two bodies performing $S.H.M.$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

Which of the following curves represents correctly the oscillation given by $y = y_0 \sin(\omega t - \phi)$,where $0 < \phi < 90^\circ$?

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If the time taken by the particle to go from $x = 0$ to $x = A/2$ is $2 \, s$,then the time taken by the particle in going from $x = A/2$ to $x = A$ is $......... \, s$.

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