The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

The displacement of a particle undergoing $SHM$ with time period $T$ is given by $x(t) = x_m \cos(\omega t + \phi)$. The particle is at $x = -x_m$ at time $t = 0$. The particle is at $x = +x_m$ when:

$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

$A$ particle executing linear $S$.$H$.$M$. has a period of $3 \ s$ and an amplitude of $6 \ cm$. The time required by it to travel a distance of $3 \ cm$ from the positive extreme position is:
$[\sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$ (in $s$)

The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

$Y = A \sin (\omega t + \phi_{0})$ 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 ...... .