Velocity and acceleration in harmonic motion

From the solution $x(t) = A\cos(\omega t + \varphi)$, differentiating once and twice, we obtain the velocity and acceleration of the oscillator:

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

The maximum values are $|v|_\text{max} = A\omega$ (reached when passing through equilibrium, where all the energy is kinetic) and $|a|_\text{max} = A\omega^2$ (reached at the extremes, where the restoring force is greatest).

The three quantities are not in phase. The velocity $v$ and the position $x$ are out of phase by $\pi/2$: when one is maximum the other is zero, and vice versa (at the point of maximum elongation the body is momentarily at rest; at the equilibrium point it moves fastest). The acceleration $a$ and the position $x$, on the other hand, are in phase opposition: a peak of $x$ corresponds to an anti-peak of $a$, consistent with $a = -\omega^2 x$.

Position (blue), velocity (gold) and acceleration (dashed red) of a harmonic oscillator. The velocity leads the position by a quarter period; the acceleration is its mirror image, flipped.

Collegamenti

Argomenti: Oscillations and harmonic motion Concetti: Simple harmonic motion Competenze: Reading graphs

Esercizi collegati: Worked exercise — Finding A and φ from initial conditions · Problem — Mass on two springs in parallel · Problem — Period and frequency of a mass-spring system