What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?

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.

Two particles $A$ and $B$ execute $SHMs$ of periods $T$ and $\frac{3T}{2}$. If they start from the mean position,then the phase difference between them,when the particle $A$ completes two oscillations,will be

Vertical displacement of a plank with a body of mass $m$ on it is varying according to the law $y = \sin \omega t + \cos \omega t$. The minimum value of $\omega$ for which the mass just breaks off the plank and the moment it occurs first after $t = 0$ are given by: ($y$ is positive vertically upwards)

The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

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