Canonical Form of Underdamped Oscillatory System

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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.


Then the value of $x$ can be expressed in the form:

$x = \dfrac {x_0 \, a} \alpha e^{-b t} \, \map \cos {\alpha t - \theta}$

where:

$\alpha = \sqrt {a^2 - b^2}$
$\theta = \map \arctan {\dfrac b \alpha}$


This can be referred to as the canonical form of the solution of $(1)$.


Proof

From Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation, the general solution of $(1)$ is:

$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$

where:

$\alpha = \sqrt {a^2 - b^2}$

This is a homogeneous linear second order ODE with constant coefficients.


Let $m_1$ and $m_2$ be the roots of the auxiliary equation:

$m^2 + 2 b + a^2 = 0$

From Solution to Quadratic Equation with Real Coefficients:

\(\displaystyle m_1\) \(=\) \(\displaystyle -b + i \sqrt {a^2 - b^2}\)
\(\displaystyle m_1, m_2\) \(=\) \(\displaystyle -b - i \sqrt {a^2 - b^2}\)

So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation:

$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$

where:

$\alpha = \sqrt {a^2 - b^2}$


The following assumptions are made:

We may label a particular point in time $t = 0$ at which the derivative of $x$ with respect to $t$ is itself zero.
We allow that at this arbitrary $t = 0$, the value of $x$ is assigned the value $x = x_0$.

This corresponds, for example, with a physical system in which the moving body is pulled from its equilibrium position and released from stationary at time zero.


Differentiating $(1)$ with respect to $t$ gives:

$\quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$


Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:

\(\displaystyle x_0\) \(=\) \(\displaystyle e^0 \paren {C_1 \cos 0 + C_2 \sin 0}\)
\(\displaystyle \) \(=\) \(\displaystyle C_1\)


Setting the initial condition $x' = 0$ when $t = 0$:

\(\displaystyle 0\) \(=\) \(\displaystyle -b e^0 \paren {C_1 \cos 0 + C_2 \sin 0} + e^0 \paren {-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}\)
\(\displaystyle \) \(=\) \(\displaystyle -b C_1 + \alpha C_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle C_2\) \(=\) \(\displaystyle \frac {b C_1} \alpha\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {b x_0} \alpha\)


Hence:

\(\displaystyle x\) \(=\) \(\displaystyle e^{-b t} \paren {x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}\)
\((2):\quad\) \(\displaystyle \) \(=\) \(\displaystyle e^{-b t} \dfrac {x_0} \alpha \paren {\alpha \cos \alpha t + b \sin \alpha t}\)


From Multiple of Sine plus Multiple of Cosine:Cosine Form, $(2)$ can be expressed as:

\(\displaystyle x\) \(=\) \(\displaystyle \dfrac {x_0} \alpha e^{-b t} \paren {\sqrt {\alpha^2 + b^2} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 \sqrt {\sqrt{a^2 - b^2}^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 \sqrt {a^2 - b^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t + \arctan \dfrac {-b} \alpha}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \arctan \dfrac b \alpha}\) Tangent Function is Odd
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \theta}\) where $\theta = \arctan \dfrac b \alpha$

$\blacksquare$


Also presented as

This can also be seen presented as:

$x = \dfrac {x_0 \sqrt {\alpha^2 + b^2} } \alpha e^{-b t} \, \cos {\alpha t - \theta}$


Sources