$A$ particle is executing simple harmonic motion with a period of $T$ seconds and amplitude $a$ meters. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}} \, m$ from its mean position in seconds is

$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)

The period of $S.H.M.$ of a particle is $16 \ s$. The phase difference between the positions at $t = 2 \ s$ and $t = 4 \ s$ will be

The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be

$A$ small mass executes linear $SHM$ about $O$ with amplitude $a$ and period $T.$ Its displacement from $O$ at time $T/8$ after passing through $O$ is:

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 $[\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}]$