Two bodies performing $SHM$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

$A$ 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 starts executing simple harmonic motion from one extreme position. If $a, b$ and $c$ are the displacements of the particle from the mean position at the ends of three successive seconds,the frequency of simple harmonic motion is

Consider a pair of identical pendulums,which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical,the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums?

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

Two particles are in $SHM$ on the same straight line with amplitudes $A$ and $2A$ and with the same angular frequency $\omega$. It is observed that when the first particle is at a distance $A/\sqrt{2}$ from the origin and moving toward the mean position,the other particle is at the extreme position on the other side of the mean position. Find the phase difference between the two particles. (in $^o$)