$A$ 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

The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

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

Explain with plots the position of a particle executing simple harmonic motion at different times.

$A$ 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:

$A$ particle executing $SHM$ of amplitude $4 \, cm$ and $T = 4 \, s$. The time taken by it to move from $+2 \, cm$ to $+2\sqrt{3} \, cm$ is