$A$ particle executes $S.H.M.$ between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is $T_1$,and from $x = A/2$ to $x = A$ is $T_2$. Then:

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

$A$ particle executes a simple harmonic motion with a periodic time $8 \text{ s}$. At time $t = 0$, it is at a mean position. The ratio of the distance traveled by a particle in the $2^{nd}$ second and that in the $1^{st}$ second of its motion is $(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1)$.

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:

Define phase at time $t$ and initial phase.