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5. Pure oscillation

The equation

\[ x'' + \omega_0^2 x = 0 \]

has characteristic roots that are purely imaginary:

\[ \lambda = \pm i\, \omega_0. \]

One option is to express the solution using complex exponentials,

\[ x(t) = c_1 e^{i\omega_0 t} + c_2 e^{-i\omega_0 t}, \]

where \(c_1\) is complex, and \(c_2=\overline{c_1}\) ensures a real-valued solution. For any complex \(z=u+iv\),

\[ z+\overline{z} = u + iv + u - iv = 2u = 2\Re(z). \]

Hence we can also write

(5.1)\[x(t) = \Re \left( c_1 e^{i\omega_0 t} \right), \qquad c_1 \in \mathbb{C},\]

having absorbed a factor of 2 into the arbitrary constant.

Suppose we define

\[ a_1 = \Re(c_1), \; a_2=-\Im(c_1) \quad \Longleftrightarrow \quad c_1 = a_1 - i a_2. \]

Then by Euler’s identity,

\[ c_1 e^{i\omega_0 t} = (a_1 - i a_2)[ \cos(\omega_0 t) + i \sin(\omega_0 t) ] = a_1 \cos(\omega_0 t) + a_2 \sin(\omega_0 t) + i\,[\cdots], \]

and the stuff in the final brackets disappears upon taking the real part. So equivalently to (5.1), we can write

(5.2)\[x(t) = a_1\cos(\omega_0 t) + a_2 \sin(\omega_0 t), \qquad a_1,a_2 \in \mathbb{R}.\]

There is an important third equivalent form.

5.1. Amplitude–phase form

Returning to (5.1), let us write \(c_1\) in polar form:

\[ c_1 = R e^{i\theta} \quad \Longleftrightarrow \quad R = |c_1|, \: e^{i \theta} = \frac{c_1}{|c_1|}. \]

Then

\[\begin{align*} x(t) & = \Re \left( c_1 e^{i\omega_0 t} \right)\\ & = \Re \left( R e^{i\theta} e^{i\omega_0 t} \right) \\ & = R\, \Re \left( e^{i(\omega_0 t+\theta)} \right) \\ & = R \cos(\omega_0 t+\theta), \end{align*}\]

where the last line follows from Euler’s identity. This is the amplitude–phase form,

(5.3)\[x(t) = R \cos(\omega_0 t+\theta), \qquad R,\theta \in \mathbb{R}.\]

A useful aspect of this form is that the solution oscillates between \(\pm R\), which is the amplitude of the oscillation.

Equations (5.1), (5.2), and (5.3) are all equivalent expressions of the general solution. We can convert between them using the identities

\[ a_1 - i a_2 = c_1 = R e^{i\theta}, \]

which amounts to simple polar/Cartesian conversions.

Example

Find the solution in amplitude-phase form of \(2x''+32x=0\) with \(x(0)=1\), \(x'(0)=-8\).

Solution

First divide through by 2 to get the standard form, \(x''+16x=0\). Then we have \(\omega_0=\sqrt{16}=4\), so

\[ x(t) = a_1 \cos(4 t) + a_2 \sin(4 t). \]

We then derive \(1=x(0)=a_1\) and \(-8=x'(0)=4a_2\), so

\[ x(t) = \cos(4 t) -2 \sin(4 t). \]

Converting \((1,-2)\) to polar form gives \(R=\sqrt{5}\) and \(\tan \theta = -2/1\). In order to find \(\theta\) numerically, you need to use a four-quadrant arctangent. In MATLAB it would be atan2(-2,1).

Example

Find the amplitude of the solution to \(x''+9x=0\), \(x(0)=-4\), \(x'(0)=6\).

Solution

The characteristic polynomial \(\lambda^2+9\) gives the roots \(\pm 3i\). We can write the solution as \(x(t) = R \cos(3 t + \theta)\) and apply the initial conditions to get

\[\begin{align*} -4 &= x(0) = R \cos(\theta), \\ 6 &= x'(0) = -3R \sin(\theta). \end{align*}\]

We are asked only for the amplitude, so there is no need to solve for \(\theta\). It’s clear that

\[ [R \cos(\theta)]^2 + [R \sin(\theta)]^2 = (-4)^2 + (-2)^2 = 20, \]

which gives \(R=\sqrt{20}=2\sqrt{5}\).

4. Complex exponentials 6. Free oscillations