Problem — Graphs of position, velocity and acceleration

Problem

Draw, on a single set of axes, the graphs of position, velocity and acceleration of a simple harmonic oscillator with amplitude $A$, angular frequency $\omega$, zero initial phase. Clearly indicate where the maxima and zeros of each quantity are, and explain in words how to derive $v$ and $a$ from the graph of $x(t)$.

Solution

With zero initial phase, position, velocity and acceleration are obtained by successive differentiation:

$\begin{aligned} x(t) &= A\cos\omega t \\ v(t) &= \dot{x} = -A\omega\sin\omega t \\ a(t) &= \ddot{x} = -A\omega^2\cos\omega t = -\omega^2 x \end{aligned}$

How to read the graph. The velocity is the slope (derivative) of the $x(t)$ curve: where $x$ passes through zero (equilibrium) its curve is steepest, so $|v|$ is maximum and equals $A\omega$; where $x$ is at the endpoints the curve is flat, so $v = 0$.

The acceleration is the slope of $v(t)$, and is also proportional to $-x$: it is maximum in magnitude at the endpoints (turning points), where it equals $A\omega^2$, and is zero at equilibrium. Hence the phase shift: $v$ leads $x$ by a quarter period, and $a$ is in phase opposition with $x$.

$\ev{v_{\max} = A\omega \text{ at the zeros of } x,\quad a_{\max} = A\omega^2 \text{ at the turning points}}$

Links

Topics: Oscillations and harmonic motion Concepts: Simple harmonic motion Skills: Graph reading