Interval between 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 $S$.

Then the successive local maxima of $x$ occur for $t = 0, T, 2T, \ldots$

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:
 * $T = \dfrac {2 \pi} {a^2 - b^2}$

Differentiating {{{WRT}} $t$:

From Derivative at Maximum or Minimum, the local maxima and local minima of $x$ occur at $x' = 0$:

It remains to be determined which of these points at which $x' = 0$ are local maxima.

This occurs when $x > 0$.

From Cosine of Angle plus Full Angle:
 * $\cos \left({x}\right) = \cos \left({2 \pi + x}\right)$

We have that at $x = x_0$ at $t = 0$.

It is given that $x_0 > 0$.

So at $t = 0, 2 \pi, 4 \pi, \ldots$ we have that:
 * $\cos \alpha t > 0$

Similarly, from Cosine of Angle plus Straight Angle:
 * $\cos \left({x}\right) = -\cos \left({\pi + x}\right)$

So at $t = \pi, 3 \pi, 5 \pi, \ldots$ we have that:
 * $\cos \alpha t < 0$

Thus we have that:
 * $\alpha L = 2 \pi$

where $L$ is the value of $t$ between consecutive local maxima of $x$.

Thus:
 * $L = \dfrac {2 \pi} {\alpha} = \dfrac {2 \pi} {a^2 - b^2} = T$

as required.