$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

$A$ particle executes $S.H.M.$ of amplitude $A$ along the $x$-axis. At $t = 0$,the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$-axis. If the displacement of the particle in time $t$ is $x = A \sin (\omega t + \delta)$,then the value of $\delta$ will be:

The minimum phase difference between two simple harmonic motions $x_1 = \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t$ and $x_2 = \sin \omega t + \cos \omega t$ is $[\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}]$

At time $t = 0$,a simple harmonic oscillator is at its extreme position. If it covers half of the amplitude distance in $1\, s$,then the time period of oscillation is ..... $s$.

Two bodies performing $SHM$ 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

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