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:

$A$ 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

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?

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?

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.

The diagram shows two oscillations. What is the phase difference between the oscillations?