Phase constant in simple harmonic motion

[Elena14]

I know the phase constant depends upon the choice of the instant t=0. Is it compulsory that the phase constant must be between [0,2π] ? I know that after 2π the motion will repeat itself so it will not really matter, but what is the conventional way to write the phase constant in the general equation of simple harmonic motion, x=A sin (wt+ φ) ; x is the displacement from the mean position, A is the amplitude, w is the angular frequency, and φ is the phase constant.

Also, when the particle starts from mean position and move towards the positive extreme, we take the phase constant to be 0 and when it moves toward the negative extreme, we take it to be π, why is that?

 

[blue_leaf77]

what is the conventional way to write the phase constant in the general equation of simple harmonic motion, x=A sin (wt+ φ)

Yes, the equation you wrote there is the conventional way to write the most general form of SHM.

lso, when the particle starts from mean position and move towards the positive extreme, we take the phase constant to be 0 and when it moves toward the negative extreme, we take it to be π, why is that?

When the particle starts from the equilibrium and takes the positive x direction in the beginning of its course, the displacement as a function of time has the form $x(t) = A \sin \omega t$, therefore $\phi = 0$. If instead, the particle drives to the negative direction at the start, the displacement will be $x(t) = -A \sin \omega t$. From there you should see that the phase constant for the second case must indeed be $\pi$.

 

[Elena14]

And how do you get x(t)=A sin ωt?
We were taught that we first calculate the phase constant and then plug it into get the general equation.

 

[blue_leaf77]

And how do you get x(t)=A sin ωt?

For the case of movement to the positive direction? Well, you need to know the boundary conditions. For example $x(0) = 0$ and $x'(0) = |v_0|$ (the initial velocity is positive because the particle moves to the positive $x$ in the beginning). Using the general solution $x(t) = A\sin (\omega t +\phi)$, the first boundary condition yields two possibilities for $\phi$: $0$ or $\pi$. The second condition entails
$$
x'(0) = |v_0| = \omega A \cos\phi
$$
The above equation will be satisfied if $\phi = 0$ instead of $\pi$, therefore
$$
x(t) = A \sin\omega t
$$
with $A = |v_0|/\omega$