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

Explain with plots the position of a particle executing simple harmonic motion at different times.

$A$ 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:

The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

Two bodies performing $SHM$ 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

$A$ simple harmonic oscillator has an amplitude $A$ and time period $6 \pi \text{ s}$. Assuming the oscillation starts from its mean position,the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x} \text{ s}$,where $x=$ . . . . . . .