Interval between Local Maxima for Underdamped Free Vibration

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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.


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:

$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$

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

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