Theorem: Interval between Local Maxima for Underdamped Free Vibration

Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
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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:
:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where:
:$\alpha = \sqrt {a^2 - b^2}$.
:$\theta = \map \arctan {\dfrac b \alpha}$
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|Differentiation}} $t$:
{{begin-eqn}}
{{eqn | l = x'
      | r = -b \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta} - \dfrac {x_0 \, a} \alpha e^{-b t} \alpha \map \sin {\omega t - \theta}
      | c = 
}}
{{eqn | r = -\dfrac {x_0 \, a} \alpha e^{-b t} \paren {b \map \cos {\alpha t - \theta} + \alpha \map \sin {\alpha t - \theta} }
      | c = 
}}
{{end-eqn}}
From the Interior Extremum Theorem, the local maxima and local minima of $x$ occur at $x' = 0$:
{{begin-eqn}}
{{eqn | l = x'
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\dfrac {x_0 \, a} \alpha e^{-b t} \paren {b \map \cos {\alpha t - \theta} + \alpha \map \sin {\alpha t - \theta} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \map \cos {\alpha t - \theta}
      | r = -\alpha \map \sin {\alpha t - \theta}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \tan {\alpha t - \theta}
      | r = -\frac b \alpha
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\tan \alpha t - \tan \theta} {1 + \tan \alpha t \tan \theta}
      | r = -\frac b \alpha
      | c = Tangent of Difference
}}
{{eqn | ll= \leadsto
      | l = \frac {\tan \alpha t - \dfrac b \alpha} {1 + \tan \alpha t \dfrac b \alpha}
      | r = -\frac b \alpha
      | c = Definition of $\theta$
}}
{{eqn | ll= \leadsto
      | l = \tan \alpha t - \dfrac b \alpha 
      | r = -\frac b \alpha - \tan \alpha t \dfrac {b^2} {\alpha^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tan \alpha t \paren {1 + \dfrac {b^2} {\alpha^2} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tan \alpha t
      | r = 0
      | c = as $1 + \dfrac {b^2} {\alpha^2} > 0$
}}
{{eqn | ll= \leadsto
      | l = \alpha t
      | r = 0, \pi, 2 \pi, \ldots
      | c = Tangent of Zero, Tangent of Angle plus Straight Angle
}}
{{end-eqn}}
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:
:$\cos x = \map \cos {2 \pi + x}$
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:
:$\cos x = -\map \cos {\pi + x}$
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.
{{qed}}