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 $\dots \; s$.

$A$ simple harmonic oscillator has an amplitude $A$ and time period $6 \pi \text{ s}$. Assuming the oscillation starts from its mean position,the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x} \text{ s}$,where $x=$ . . . . . . .

The phase of a particle executing simple harmonic motion is $\frac{\pi}{2}$ when it has:

The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

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

$A$ particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.