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

After the beginning of motion,how long will it take for a harmonically oscillating particle to reach a displacement equal to one-half of its amplitude,if the time period is $24 \ sec$ and the particle starts from rest?

$A$ particle is executing $SHM$ of amplitude $A$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +A/2$ and $x = -A/\sqrt{2}$ (in $^o$)?

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.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $x = \frac{A}{2}$ is $2 \text{ s}$, then time taken by particle in going from $x = \frac{A}{2}$ to $x = A$ is (in $\text{ s}$)

Which of the following statements is correct for $S.H.M.$?