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

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

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

$A$ particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ and travels towards the left as shown in the figure. If the time period of $SHM$ executed about $O$ is $T$ and amplitude is $A$,then the equation of motion of the particle is:

$A$ 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 $x=$ . . . . . . .

$A$ particle executes $S.H.M.$ between $x = -A$ and $x = +A$. The time taken for it to go from $x = 0$ to $x = A/2$ is $T_1$,and from $x = A/2$ to $x = A$ is $T_2$. Then: