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

$A$ particle performs simple harmonic motion with a period of $3 \ s$. The time taken by it to cover a distance equal to half the amplitude from the mean position is $\left[\sin 30^{\circ}=0.5\right]$.

$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

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

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

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