$A$ spring with a spring constant $1200 \; N m^{-1}$ is mounted on a horizontal table as shown in the figure. $A$ mass of $3 \; kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0 \; cm$ and released. Let us take the position of the mass when the spring is unstretched as $x = 0$,and the direction from left to right as the positive direction of the $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch $(t = 0)$,the mass is:
$(a)$ at the mean position,
$(b)$ at the maximum stretched position,and
$(c)$ at the maximum compressed position.
In what way do these functions for $SHM$ differ from each other: in frequency,in amplitude,or in the initial phase?

(N/A) The functions have the same frequency and amplitude,but different initial phases.
Amplitude of oscillation,$A = 2.0 \; cm = 0.02 \; m$.
Force constant of the spring,$k = 1200 \; N m^{-1}$.
Mass,$m = 3 \; kg$.
Angular frequency of oscillation,$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = \sqrt{400} = 20 \; rad s^{-1}$.
$(a)$ When the mass is at the mean position at $t = 0$,the initial phase is $0$. The displacement is $x = A \sin(\omega t) = 0.02 \sin(20t)$.
$(b)$ At the maximum stretched position (extreme right),the initial phase is $\frac{\pi}{2}$. The displacement is $x = A \sin(\omega t + \frac{\pi}{2}) = A \cos(\omega t) = 0.02 \cos(20t)$.
$(c)$ At the maximum compressed position (extreme left),the initial phase is $\frac{3\pi}{2}$ (or $-\frac{\pi}{2}$). The displacement is $x = A \sin(\omega t + \frac{3\pi}{2}) = -A \cos(\omega t) = -0.02 \cos(20t)$.
These functions for $SHM$ differ from each other only in their initial phases.