$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

$A$ 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

$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ 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?

Two particles executing $S.H.M.$ of same frequency,meet at $x=+A/2$,while moving in opposite directions. Phase difference between the particles is .........

The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?