$A$ particle performs simple harmonic motion with a period of $3 \ s$. The time taken by it to cover a distance equal to half the amplitude from the mean position is $\left[\sin 30^{\circ}=0.5\right]$.

$A$ particle executes simple harmonic motion. Its amplitude is $8 \,cm$ and time period is $6 \,s$. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude,is ............. $s$

$A$ particle executing linear $S$.$H$.$M$. has a period of $3 \ s$ and an amplitude of $6 \ cm$. The time required by it to travel a distance of $3 \ cm$ from the positive extreme position is:
$[\sin 30^{\circ} = \cos 60^{\circ} = \frac{1}{2}, \sin 60^{\circ} = \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$ (in $s$)

If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?

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

The displacement versus time curve for a particle executing $SHM$ is shown in the figure. Identify the points marked at which $(i)$ the velocity of the oscillator is zero,$(ii)$ the speed of the oscillator is maximum.