Mechanical analogy — spring and LC

The equation $\ddot{Q} = -(1/LC)\,Q$ is identical to $\ddot{x} = -(k/m)\,x$ for a spring: the LC circuit is the electrical analogue of a harmonic oscillator. The correspondence is so precise that every quantity can be translated term by term from one world to the other.

Mechanical (spring)	Electrical (LC)
position $x$	charge $Q$
velocity $\dot{x}$	current $i$
mass $m$	inductance $L$
spring constant $k$	inverse capacitance $1/C$
elastic energy $\tfrac{1}{2}kx^2$	electrical energy $\tfrac{1}{2}Q^2/C$
kinetic energy $\tfrac{1}{2}m\dot{x}^2$	magnetic energy $\tfrac{1}{2}Li^2$
$\omega = \sqrt{k/m}$	$\omega = 1/\sqrt{LC}$

The dictionary clarifies the physical role of each component. Inductance plays the role of mass: it is the inertia of the circuit, what resists changes in current (just as mass resists changes in velocity). The inverse of capacitance plays the role of stiffness: a small capacitor (large $1/C$) is like a stiff spring, it pulls the charge back strongly and makes it oscillate faster.

$\ddot{Q} = -\frac{1}{LC}\,Q \qquad\longleftrightarrow\qquad \ddot{x} = -\frac{k}{m}\,x$

Collegamenti

Argomenti: Electromagnetic induction Concetti: LC and RLC oscillations · Simple harmonic motion

Esercizi collegati: Worked exercise — LC circuit, find the natural frequency · Problem — Natural frequency of an LC circuit · Problem — Capacitor of an AM radio receiver