The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is

$A$ particle executes simple harmonic motion represented by the displacement function $x(t) = A \sin (\omega t + \phi)$. If the position and velocity of the particle at $t = 0 \, s$ are $2 \, cm$ and $2 \omega \, cm \, s^{-1}$ respectively,then its amplitude is $x \sqrt{2} \, cm$,where the value of $x$ is ..... .

$A$ 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:

$A$ particle oscillates in a straight line simple harmonically with a period of $8 \ s$ and an amplitude of $4 \sqrt{2} \ m$. The particle starts from the mean position. The ratio of the distance travelled by it in the $1^{\text{st}}$ second of its motion to that in the $2^{\text{nd}}$ second is: $(\sin 45^{\circ} = 1 / \sqrt{2}, \sin \frac{\pi}{2} = 1)$

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

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is: