# What is the difference between amplitude and phase angle for a pendulum?

I understand the definition in terms a plain trigonometric function, such as $$A\cos(\omega t+\phi)$$. The amplitude is half of the distance between peak and bottom, and the phase angle* is how the function shifts horizontally.

However, I am having a hard time grasp how would these two terms apply to a real-world pendulum. If a pendulum's amplitude is the maximum displacement from its equilibrium position, then what is its phase angle? Is it the initial displacement? If so, wouldn't the phase angle already determine the amplitude, if we simply release the pendulum without applying any extra force?

Look at a solution of the (linearized) equation of motion for the pendulum, $$\theta(t)=A\cos\left(\sqrt{\frac{g}{l}}t+\varphi\right).$$ The amplitude $$|A|$$ is the maximum value of $$\theta$$; the angle varies between $$-|A|$$ and $$|A|$$ in the course of each period of length $$T=2\pi\sqrt{l/g}$$.
The phase angle $$\varphi$$ tells you where in the cycle the pendulum is at any particular time. For example, if you start with the pendulum lifted up to an angle $$\theta_{0}$$ and you release it at time $$t=0$$, the solution will have $$A=\theta_{0}$$ and $$\varphi=0$$, since the initial conditions have $$\theta(0)$$ at a maximum and the angular velocity $$\dot{\theta}(0)=0$$. If instead you start the pendulum at its equilibrium position at $$t=0$$, but with a nonzero angular velocity $$\dot{\theta}(0)>0$$, the phase angle will be $$\varphi=-\frac{\pi}{2}$$ (or $$-90^{\circ}$$), making the solution $$\theta(t)=A\cos\left(\sqrt{\frac{g}{l}}t-\frac{\pi}{2}\right)= A\sin\left(\sqrt{\frac{g}{l}}t\right).$$