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

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 $^{\circ}$)

The phase (at a time $t$) of a particle in simple harmonic motion tells:

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. If time taken by particle to go from $x = 0$ to $x = \frac{A}{2}$ is $2 \text{ s}$, then time taken by particle in going from $x = \frac{A}{2}$ to $x = A$ is (in $\text{ s}$)

Which of the following statements is correct for $S.H.M.$?

$A$ particle performs simple harmonic motion between $x = -A$ and $x = +A$. How much time does it take to travel from $x = A$ to $x = A/2$?