Canonical Form of Underdamped Oscillatory System

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:

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 $x$ $t$ 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 $(1)$ $t$ 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 $(1)$:

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

Hence:

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

Also presented as
This can also be seen presented as:


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