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

(N/A) The figure shows the positions of a particle executing $S.H.M.$ at discrete time intervals,where each interval is $\frac{T}{4}$,assuming initial phase $\phi=0$ and $T$ is the period of motion.
For a given $S.H.M.$,if $A$ is the amplitude,the position of a particle at time $t$ is determined by the phase $(\omega t+\phi)$ of the cosine function.
The general equation of $S.H.M.$ is:
$x(t) = A \cos(\omega t + \phi)$
Given $\phi = 0$ and $\omega = \frac{2\pi}{T}$,the equation becomes:
$x(t) = A \cos\left(\frac{2\pi}{T} t\right)$
Now,we can determine the position at different times:
$1$. At $t = 0$: $x(0) = A \cos(0) = +A$
$2$. At $t = \frac{T}{4}$: $x\left(\frac{T}{4}\right) = A \cos\left(\frac{2\pi}{T} \cdot \frac{T}{4}\right) = A \cos\left(\frac{\pi}{2}\right) = 0$
$3$. At $t = \frac{T}{2}$: $x\left(\frac{T}{2}\right) = A \cos\left(\frac{2\pi}{T} \cdot \frac{T}{2}\right) = A \cos(\pi) = -A$
$4$. At $t = \frac{3T}{4}$: $x\left(\frac{3T}{4}\right) = A \cos\left(\frac{2\pi}{T} \cdot \frac{3T}{4}\right) = A \cos\left(\frac{3\pi}{2}\right) = 0$
$5$. At $t = T$: $x(T) = A \cos\left(\frac{2\pi}{T} \cdot T\right) = A \cos(2\pi) = +A$