$A$ 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

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

$A$ 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 $(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}, \sin 90^\circ = \cos 0^\circ = 1)$.

$A$ large number of random snapshots using a camera are taken of a particle in simple harmonic motion between $x = -x_0$ and $x = +x_0$,with the origin $x = 0$ as the mean position. $A$ histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble:

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 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: