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PAGE 1

9.4 Application of Fourier Series

\[ m x'' + c x' + k x = F(t) \]

mass (coefficient of \( x'' \))

damping constant (coefficient of \( x' \))

spring constant (coefficient of \( x \))

Solution Structure

\[ x(t) = x_c(t) + x_p(t) \]
  • \( x_c(t) \) (complementary): The solution with the right side equal to 0.
  • \( x_p(t) \) (particular): The solution due to the input \( F(t) \).

Case: Sinusoidal Input

If \( F(t) = F_0 \sin(\omega t) \), we use the method of undetermined coefficients:

\[ x_p = A \cos(\omega t) + B \sin(\omega t) \quad \text{if } \omega \neq \sqrt{\frac{k}{m}} \]

Case: Periodic Input

If \( F(t) \) is periodic, break it into cosines and sines using a Fourier series.

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Example: Periodic Forcing Function

Given: \( c = 0, m = 1, k = 5 \)

\[ F(t) = \begin{cases} 1 & 0 < t < 3 \\ -1 & 3 < t < 6 \end{cases} \quad \text{period } 6 \]
Graph of a square wave function F(t) with period 6, showing origin symmetry.

Fourier Series Expansion

Due to origin symmetry, \( F(t) \) is a sine series:

\[ a_n = 0, \quad b_n = \frac{2}{L} \int_{0}^{L} F(t) \sin\left(\frac{n \pi t}{L}\right) dt \]

Calculating the coefficients for \( L = 3 \):

\[ F(t) = \sum_{n=1}^{\infty} \frac{2}{n \pi} [1 - (-1)^n] \sin\left(\frac{n \pi t}{3}\right) \]
\[ F(t) = \frac{4}{\pi} \sin\left(\frac{\pi t}{3}\right) + \frac{4}{3\pi} \sin\left(\frac{3\pi t}{3}\right) + \frac{4}{5\pi} \sin\left(\frac{5\pi t}{3}\right) + \dots \]

Differential Equation

\[ x'' + 5x = F(t) \]

Each term in the Fourier series of \( F(t) \) results in a particular solution.

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Particular and General Solutions for Differential Equations

\[x_p = A_n \cos\left(\frac{n\pi t}{3}\right) + B_n \sin\left(\frac{n\pi t}{3}\right) \quad n=1, 2, 3, \dots\]

the whole particular solution is

\[x_p = \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi t}{3}\right) + B_n \sin\left(\frac{n\pi t}{3}\right)\]

that \(x_p\) must satisfy

\[x'' + 5x = F(t) = \sum_{n=1}^{\infty} \frac{2}{n\pi} \left[1 - (-1)^n\right] \sin\left(\frac{n\pi t}{3}\right)\]

sub \(x_p\) into the equation above:

\(A_n = 0\) for all \(n\)

\[B_n = \frac{18 \left[1 - (-1)^n\right]}{n\pi (45 - n^2\pi^2)}\]

General Solution:

\[x(t) = \underbrace{C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t)}_{x_c} + \underbrace{\sum_{n=1}^{\infty} \frac{18 \left[1 - (-1)^n\right]}{n\pi (45 - n^2\pi^2)} \sin\left(\frac{n\pi t}{3}\right)}_{x_p}\]

\(x_p\) is also the steady-periodic solution.

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Analysis of Resonance

\(45 - n^2\pi^2 \neq 0\) for \(n = 1, 2, 3, \dots\)

close to 0 w/ \(n=2\) because it nearly matches the natural frequency \(\sqrt{5} \approx 2.236\)

input freq \(\frac{n\pi}{3} = \frac{2\pi}{3} \approx 2.094\)

(near resonance)

for this one, it's ok because if \(n=2\), \(B_n = 0\) because \([1 - (-1)^n]\)

Alternative Case

if we had

\[x'' + 5x = \begin{cases} 1 & 0 < t < 10 \\ -1 & 10 < t < 20 \end{cases} \quad \text{period } 20\]
\[x_p = \sum_{n=1}^{\infty} \frac{200 \left[1 - (-1)^n\right]}{n\pi (500 - n^2\pi^2)} \sin\left(\frac{n\pi t}{10}\right)\]

\(500 - n^2\pi^2\) is close to 0 w/ \(n=7\)

\([1 - (-1)^n]\) does NOT eliminate it

near resonance is visible

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Mass-Spring Response to \( F(t) = \pm 1 \) (\( x'' + 5x = F(t) \))

Plot of amplitude vs time showing square wave forcing and resulting oscillatory responses.

Legend

  • Forcing \( F(t) \) (Square Wave)
  • Homogeneous \( x_c(t) \) (Natural Response)
  • Steady Periodic \( x_p(t) \) (Forced Response)
  • General Solution \( x(t) \) (Total)

The graph illustrates the interaction between the natural response of the system and the forced response driven by a square wave input. The total solution \( x(t) \) is the sum of the homogeneous and steady periodic components.

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Mass-Spring Response with \( T = 20 \): Visualizing Near-Resonance

Dominant \( n = 7 \) harmonic due to near-resonance (\( \omega_7 \approx \omega_0 \))

Graph of amplitude vs time showing high-frequency oscillations and near-resonance effects.

Legend

  • Forcing \( F(t) \) (Input)
  • Homogeneous \( x_c(t) \) (Natural Ringing)
  • Steady Periodic \( x_p(t) \) (Forced Response)
  • General Solution \( x(t) \) (Total Motion)

This visualization highlights the phenomenon of near-resonance, where a specific harmonic of the forcing function (in this case, the 7th harmonic) closely matches the natural frequency of the system, leading to amplified oscillations.

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Analysis of Pure Resonance in Differential Equations

Now we look at:

\[ x'' + 9x = \begin{cases} 1 & 0 < t < \pi \\ -1 & \pi < t < 2\pi \end{cases} \quad \text{period } 2\pi \]
\(\vdots\)
\[ x_p = \sum_{n=1}^{\infty} \frac{2(1 - (-1)^n)}{n(9 - n^2)} \sin(nt) \]

\( n = 3 \) makes \( 9 - n^2 = 0 \) pure resonance

The series solution \( x_p \) is not valid for \( n = 3 \).

Consider that one term separately:

complementary: \( x'' + 9x \) = \( A t \cos(3t) + B t \sin(3t) \)
\( \cos(3t) \)
\( \sin(3t) \)
\( x_p \)
\(\vdots\)
\( A = -\frac{2}{3} \) \( \quad B = 0 \)
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Pure Resonance vs. Damped Response: \( x'' + cx' + 9x = F(t) \)

Plot of Amplitude vs Time showing blue oscillating waves growing linearly and orange waves leveling off.

Key Observations from the Plot:

  • Undamped (c = 0): Shows pure resonance with linear growth of amplitude over time.
  • Damped (c = 0.5): Reaches a steady state where the amplitude levels off due to damping.
  • Forcing F(t): Represented by the dashed square wave background.
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General Solution

\[ x(t) = C_1 \cos(3t) + C_2 \sin(3t) - \frac{2}{3} t \cos(3t) + \sum_{\substack{n=1 \\ n \neq 3}}^{\infty} \frac{2 [1 - (-1)^n]}{n(9 - n^2)} \sin(nt) \]

Damped Case: If \( c \neq 0 \)

Consider the differential equation: \[ mx'' + cx' + kx = F(t) \]

Where the forcing function is given by the Fourier series: \[ F(t) = \sum_{n=1}^{\infty} F_n \sin\left( \frac{n \pi t}{L} \right) \]

The particular solution \( x_p \) is: \[ x_p = \sum_{n=1}^{\infty} \frac{F_n}{\sqrt{(k - m \omega_n^2)^2 + (c \omega_n)^2}} \sin\left( \frac{n \pi t}{L} - \tan^{-1}\left( \frac{c \omega_n}{k - m \omega_n^2} \right) \right) \]

\[ \omega_n = \frac{n \pi}{L} \]

if \( c \neq 0 \)

denom \( \neq 0 \)

damper introduces a phase shift

(steady periodic lags complementary)

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Damped Response at Resonance

Equation: \( x'' + 0.5x' + 9x = F(t) \) where \( c = 0.5 \)

A line graph titled 'Damped Response at Resonance' plotting Amplitude from -1.00 to 1.00 against Time (t) from 0 to 40. It shows four curves: a square-wave forcing function F(t), a decaying homogeneous transient x_c(t), a steady periodic x_p(t), and the total general solution x(t) which eventually synchronizes with the steady periodic curve.

Transient "ringing" \( (x_c) \) decays due to damping.

Total motion \( (x) \) converges to Steady Periodic \( (x_p) \).