$A$ 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:

The displacement of a particle executing simple harmonic motion is $y = A \sin (2t + \phi) \ m$,where $t$ is time in seconds and $\phi$ is the phase angle. At time $t = 0$,the displacement and velocity of the particle are $2 \ m$ and $4 \ ms^{-1}$ respectively. The phase angle $\phi$ is: (in $^{\circ}$)

$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

$A$ particle is executing $SHM$. The time taken for $\left(\frac{3}{8}\right)^{\text{th}}$ of an oscillation from extreme positions is $x$. Then,the time taken for the particle to complete $\left(\frac{5}{8}\right)^{\text{th}}$ of an oscillation from the mean position 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?

$A$ 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