We just began a new topic on oscillation and simple harmonic motion. I'm having quite a hard time grasping what the purpose of the phase constant that appears in the argument of the cosine function. Reading online, I found that it has something to do with either shifting the graph, or with the position of oscillation at $t=0$.
[Physics] Phase constant in simple harmonic motion
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Best Answer
We can characterise harmonic motion with $x(t) = A\cos(\omega t + \phi)$ for displacement $x,$ amplitude $A,$ angular frequency $\omega$ and phase constant $\phi$.
At $t=0$ when the oscillation starts, we get $x(0) = A\cos(\phi)$. If $\phi = 0$ then we simply get $x(0) = A.$ As in the motion starts at the maximum amplitude.
However if we have the motion starting at the centre of oscillation, with some negative velocity that would mean $x(0) = 0.$ This means $\cos(\phi) = 0$ and so $\phi = \pi/2$ (or $3\pi/2$, but think about what that would mean for the velocity).
Essentially the phase constant $\phi$ determines the initial position of the oscillation, at $t=0.$ As $\phi$ goes from $0$ to $2\pi$, the initial position goes from $A$ to $-A$ and back to $A$, as the cosine of the phase.