Time period of a particle executing $SHM$ is $8 \, s$. At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is:

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

$A$ particle executes a simple harmonic motion with a periodic time $8 \text{ s}$. At time $t = 0$, it is at a mean position. The ratio of the distance traveled by a particle in the $2^{nd}$ second and that in the $1^{st}$ second of its motion is $(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1)$.

$A$ particle executes simple harmonic motion (amplitude $= A$) between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is ${T_1}$ and to go from $x = A/2$ to $x = A$ is ${T_2}$. Then:

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

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