$A$ 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 is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

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.

$A$ 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 $s$)

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?

Explain what is phase and draw a single graph showing different phases of simple harmonic motion.