Phase constant in simple harmonic motion 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 $t=0$.
 A: 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.
A: I'd like to add some more general remarks to the perfectly fine answer by Nuclear_Wizard.
Let's say your harmonic oscillator is characterized by the equation:
$$m\ddot{x}+kx=0$$
You can see from this equation that it is time translation invariant, meaning it does not change if you shift time from $t$ to $t'=t+\tau$. This expresses the fact that the motion of a harmonic oscillator is independent of "absolute time", i.e. it does not matter whether you run your oscillator today, tomorrow or in a week, the motion will always be the same and given by the above equation.
Now, looking at the general solution for this differential equation:
$$x(t)=A\cos(\omega t+\phi)=A\cos\left(\omega (t+\frac{\phi}{\omega})\right),$$
where $x$ is displacement, $A$ is the amplitude, $\omega=\sqrt{k/m}$ the angular frequency and $\phi$ the phase. This time translation invariance reappears here in the form of an arbitrary constant $\phi$ which acts as an arbitrary shift in time (by $\phi/\omega$).
Mathematically, the general solution of a second order ordinary differential equation is given up to two constants (in this case $\phi$ and $A$).
In order to describe a concrete oscillator (experiment) you will typically specify its position $x$ and velocity $\dot{x}$ at some moment in time, e.g. at $t=0$ (other combinations are possible, e.g. position at $t=0$ and $t=1~\mathrm{min}$,...). Technically this is called an initial value problem for the ODE of the oscillator. These conditions will fix the phase $\phi$ and amplitude, $A$. 

Note that this is not limited to the motion of an oscillator. For instance for the free fall with gravitational acceleration $g$:
$$\ddot{y}=-g$$
you have the general solution:
$$y=-\frac{gt^2}{2}+v_0 t+y_0,$$
You can rewrite this equation as:
$$y=-g\left(t-\frac{v_0}{2g}\right)^2+y_0+\frac{v_0^2}{4g}$$
Similarly to the oscillator you have the time shifted by an arbitrary constant. Boundary or initial conditions will fix your constants such that $v_0$ is the initial velocity at $t=0$ and $y_0$ is the initial height at $t=0$. 
However the term phase is specific to oscillatory motion. You would not call the constants for the free fall "phase".
