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:
- $(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
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 {with respect to $t$:
\(\ds x'\) | \(=\) | \(\ds -b \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta} - \dfrac {x_0 \, a} \alpha e^{-b t} \alpha \map \sin {\omega t - \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {x_0 \, a} \alpha e^{-b t} \paren {b \map \cos {\alpha t - \theta} + \alpha \map \sin {\alpha t - \theta} }\) |
From Derivative at Maximum or Minimum, the local maxima and local minima of $x$ occur at $x' = 0$:
\(\ds x'\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -\dfrac {x_0 \, a} \alpha e^{-b t} \paren {b \map \cos {\alpha t - \theta} + \alpha \map \sin {\alpha t - \theta} }\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \map \cos {\alpha t - \theta}\) | \(=\) | \(\ds -\alpha \map \sin {\alpha t - \theta}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \tan {\alpha t - \theta}\) | \(=\) | \(\ds -\frac b \alpha\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\tan \alpha t - \tan \theta} {1 + \tan \alpha t \tan \theta}\) | \(=\) | \(\ds -\frac b \alpha\) | Tangent of Difference | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\tan \alpha t - \dfrac b \alpha} {1 + \tan \alpha t \dfrac b \alpha}\) | \(=\) | \(\ds -\frac b \alpha\) | Definition of $\theta$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tan \alpha t - \dfrac b \alpha\) | \(=\) | \(\ds -\frac b \alpha - \tan \alpha t \dfrac {b^2} {\alpha^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tan \alpha t \paren {1 + \dfrac {b^2} {\alpha^2} }\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tan \alpha t\) | \(=\) | \(\ds 0\) | as $1 + \dfrac {b^2} {\alpha^2} > 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \alpha t\) | \(=\) | \(\ds 0, \pi, 2 \pi, \ldots\) | Tangent of Zero, Tangent of Angle plus Straight Angle |
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.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Problem $(2)$