I know the phase constant depends upon the choice of the instant $t=0$. Is it compulsory that the phase constant must be between $[0,2 \pi]$? I know that after $2\pi$ 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, $x=A \sin (\omega t+ \phi)$; $x$ is the displacement from the mean position, $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ 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 $\pi$, why is that?
Best Answer
In the simplest example the concept of phase angle is a convenient way of comparing the motion of two simple harmonic oscillations of the same frequency. $\omega = \frac {2 \pi}{T}$ where $T$ is the period of the oscillation.
If you have two oscillations an oscillation $A$ has a maximum displacement at time $t_A$ and oscillation $B$ reaches a maximum displacement at a time $t_B$ then the phase angle $\phi_{BA}$ can be said to be $ \dfrac {t_B-t_A}{T} \cdot 2 \pi$ where $T$ is the period of the motion. This is the phase of $B$ relative to $A$.
However one could equally say that the phase of $A$ relative to $B$ is what is required then $\phi_{AB} = -\phi_{BA}$.
But then since the motion is continuous you can have other time differences when the motion $A$ is at the same relative position to that of the motion $B$.
The one could write $ \phi_{BA} =\dfrac {(t_B+nT)-t_A}{T} \cdot 2 \pi = \left (\dfrac {t_B-t_A}{T} + n\right)\cdot 2 \pi$ where $n$ is an integer. This represents motion $B$ being many periods and a little bit behind that of motion $A$.
For convenience the phase angle is restricted to the ranges $0\le \phi \le \pi$ or $-\frac \pi 2 \le \phi \le +\frac \pi 2$
The same sort of analysis is true for the motions at different positions and then the period $T$ would be replaced by the wavelength $\lambda$.