Ratio of Successive Local Maxima for Underdamped Free Vibration

Theorem
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
 * $(1): \quad \dfrac {\mathrm d^2 x} {\mathrm d t^2} + 2 b \dfrac {\mathrm d x} {\mathrm d t} + a^2 x = 0$

for $a, b \in \R_{>0}$.

Let $b < a$, so as to make $S$ underdamped.


 * UnderdampedPeriodAmplitude.png

Let $T$ be the period of oscillation of $S$.

Let $x_1$ and $x_2$ be successive local maxima of $x$.

Then:
 * $\dfrac {x_1} {x_2} = e^{b T}$

Proof
Let the position of $S$ be described in the canonical form:
 * $(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right)$

where:
 * $\alpha = \sqrt {a^2 - b^2}$.
 * $\theta = \arctan \left({\dfrac b \alpha}\right)$

From Period of Oscillation of Underdamped System is Regular, the period of oscillation $T$ is given by:
 * $T = \dfrac {2 \pi} {a^2 - b^2}$

From Interval between Local Maxima for Underdamped Free Vibration, the local maxima of $x$ occur at:
 * $t = 0, T, 2 T, \ldots$

Let $x_1$ occur at $n T$.

Thus $x_2$ occurs at $\left({n + 1}\right) T$.

Differentiating {{{WRT}} $t$:

As $T = \dfrac {2 \pi} {a^2 - b^2} = \dfrac {2 \pi} \alpha$, we have:

Hence:

as required.