We know that for a simple harmonic linear oscillator, the displacement is given by $x(t)=A\sin(\omega t + \phi)$, where $\phi$ denotes the phase angle. Now as per my understanding this $\phi$ is only significant when considering SHM in form of a sinusoidal wave. Is there any physical meaning in reality. Is there a way to measure the phase angle in reality just by virtue of the particle's (which is oscillating) position with respect to the mean position?
Some Assumptions
- I am considering SHM in only one axis.
- I am also considering SHM linearly for ex Spring Block System.
Best Answer
The mean position of what you wrote above is zero.
The phase angle $\phi$ (along with the other parameters $A$ and $\omega$) tells you the initial position and velocity.
You can re-write $x(t)$ as $$ x(t) = A\left(\sin(\omega t)\cos(\phi)+\cos(\omega t)\sin(\phi)\right)\;, $$ to see that the initial position is given by $$ x(0) = A\sin(\phi)\;, $$ and the initial velocity is given by $$ v(0) = A\omega\cos(\phi)\;. $$
You can solve for $\phi$ like: $$ \phi = \tan^{-1}\left(\frac{\omega x(0)}{v(0)}\right) $$