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PAGE 1

3.7 (continued)

Last time: \( mu'' + ku = 0 \)

Example: \( m = \frac{5}{16} \), \( k = 20 \)

\( u(0) = \frac{1}{6} \), \( u'(0) = -1 \) (upward)

Solved: \( u(t) = \frac{1}{6} \cos(8t) - \frac{1}{8} \sin(8t) \)

Converting Form

Want to turn \( u(t) = A \cos(\omega_0 t) + B \sin(\omega_0 t) \) into \( u(t) = R \cos(\omega_0 t - \delta) \)

Identity: \( \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b) \)

\( u(t) = R \cos(\delta)\cos(\omega_0 t) + R \sin(\delta)\sin(\omega_0 t) \)

We see:

\( A = R \cos(\delta) \)

\( B = R \sin(\delta) \)

\( R = \sqrt{A^2 + B^2} \)

\( \tan(\delta) = \frac{B}{A} \)

\( A^2 + B^2 = R^2 \cos^2(\delta) + R^2 \sin^2(\delta) \)

\( A^2 + B^2 = R^2 \)

Diagram of a mass m attached to a spring with constant k, showing elongation u from equilibrium where u > 0 is down.
0 is down.">
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\( u(t) = \frac{1}{6} \cos(8t) - \frac{1}{8} \sin(8t) \)

\( u(t) = \frac{5}{24} \cos(8t - (-0.64)) \)

\( R = \sqrt{(\frac{1}{6})^2 + (-\frac{1}{8})^2} = \frac{5}{24} \)

\( \tan(\delta) = \frac{-1/8}{1/6} = -\frac{3}{4} \)

\( \delta = \tan^{-1}(-\frac{3}{4}) \approx -0.64 \text{ rad} \)

Parameters

  • Amplitude (R): \( \frac{5}{24} \) - how far up and down cosine goes.
  • Phase shift (\( \delta \)): \( -0.64 \) - how the peak of cosine is moved (\( \delta > 0 \rightarrow \) move RIGHT).
  • Angular frequency (\( \omega_0 \)): 8 - how fast the oscillation is.
  • Period: \( T = \frac{2\pi}{\omega_0} \). Here \( T = \frac{2\pi}{8} = \frac{\pi}{4} \).
Graph of u(t) vs t showing a cosine wave with amplitude 5/24, period T, and first peak at delta = -0.64.
PAGE 3

Displacement from Equilibrium

The following graph represents the displacement function \( u(t) \), where positive displacement is defined as being in the downward direction.

\[ u(t) = R \cos(\omega_0 t - \delta) = \frac{5}{25} \cos(8t + 0.64) \]
Graph of u(t) vs t showing a red cosine wave with labeled amplitude R, phase shift delta, and period T.

Key Parameters

  • Amplitude (\( R \)): The maximum displacement from the equilibrium position.
  • Phase Shift (\( \delta \)): The horizontal shift of the wave from the origin.
  • Period (\( T \)): The time required to complete one full cycle, defined as:
    \[ T = \frac{2\pi}{\omega_0} \]
PAGE 4

Damped Harmonic Motion

Now let's add a damper to the system.

Schematic of a mass m suspended by a spring with constant k and a damper with constant gamma.

Damping Properties

The system parameters are defined as \( (k > 0, \gamma > 0, m > 0) \).

  • \( \gamma \): damping constant
  • Resists velocity (\( u' \)) in the opposite direction
  • Force \( = -\gamma u' \) (opposite direction)
  • Just like spring force \( = -ku \)

New Equation of Motion

\[ mu'' + \gamma u' + ku = 0 \]

(\( u > 0 \) is down)

Initial conditions: \( u(0) = u_0, u'(0) = v_0 \)

Characteristic Equation

\[ mr^2 + \gamma r + k = 0 \] \[ r = \frac{-\gamma \pm \sqrt{\gamma^2 - 4km}}{2m} \]

\( \gamma^2 - 4km \) determines the solution type.

PAGE 5

Damping Classifications

if \( \gamma^2 - 4km < 0 \) \( \rightarrow \) \( r \)'s are complex

Solutions are decaying cosines and sines

has oscillations

underdamped

"weak" damping

if \( \gamma^2 - 4km = 0 \) \( \rightarrow \) \( r \)'s are real and repeated

Solutions are decaying exponentials. No oscillations.

critically damped

damped "just right"

if \( \gamma^2 - 4km > 0 \) \( \rightarrow \) \( r \)'s are real, distinct, negative

Solutions are also decaying exponentials.

No oscillations

overdamped

"strong" damping

PAGE 6

Example: Adding a Damper

Back to the example we did last time, add a damper:

\[ mu'' + \gamma u' + ku = 0 \]
\[ m = \frac{5}{16}, \quad k = 20, \quad \gamma = 4 \]\[ u(0) = \frac{1}{6}, \quad u'(0) = -1 \]

\( \gamma^2 < 4km \)

underdamped

\[ \frac{5}{16}u'' + 4u' + 20u = 0 \]
\[ r = \frac{-4 \pm \sqrt{16 - 4 \cdot \frac{5}{16} \cdot 20}}{2(\frac{5}{16})} = \frac{-4 \pm \sqrt{-9}}{\frac{5}{8}} = -\frac{32}{5} \pm \frac{24}{5}i \]
\[ u(t) = c_1 e^{-\frac{32}{5}t} \cos\left(\frac{24}{5}t\right) + c_2 e^{-\frac{32}{5}t} \sin\left(\frac{24}{5}t\right) \]

Using \( u(0) = \frac{1}{6} \), \( u'(0) = -1 \), we get:

\[ u(t) = \frac{1}{6} e^{-\frac{32}{5}t} \cos\left(\frac{24}{5}t\right) + \frac{1}{72} e^{-\frac{32}{5}t} \sin\left(\frac{24}{5}t\right) \]\[ = e^{-\frac{32}{5}t} \left[ \frac{1}{6} \cos\left(\frac{24}{5}t\right) + \frac{1}{72} \sin\left(\frac{24}{5}t\right) \right] \]

turn into \( R \cos(\omega t - \delta) \)

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\[ = e^{-\frac{32}{5}t} \left[ \frac{\sqrt{145}}{72} \cos \left( \frac{24}{5}t - \tan^{-1}(\frac{1}{12}) \right) \right] \]
\[ = \frac{\sqrt{145}}{72} e^{-\frac{32}{5}t} \left[ \cos \left( \frac{24}{5}t - 0.08 \right) \right] \]

Amplitude

goes to 0 as \( t \to \infty \)

Quasi frequency

(damped freq)

Phase shift

\( -0.08 \)

Quasi period

\( T = \frac{2\pi}{\omega} \)

“quasi” because we never return to the same \( u \) due to damping

Graph of a damped oscillation u(t) vs t, bounded by exponential decay envelopes \pm \frac{\sqrt{145}}{72} e^{-\frac{32}{5}t} .
PAGE 8

Comparison of Undamped and Damped Motion

undamped: \( \frac{5}{24} \cos(8t + 0.64) \)

damped: \( \frac{\sqrt{145}}{72} e^{-32/5 t} (\cos \frac{24}{5}t - 0.08) \)

frequency:

\( 8 \to \frac{24}{5} \)

damper decreased the frequency

period:

\( \frac{2\pi}{8} \to \frac{2\pi}{24/5} \)

period increased

PAGE 9

Damping Behavior in Oscillatory Systems

The following graph illustrates the displacement \( x(t) \) over time \( t \) for different damping conditions in a harmonic oscillator. The curves represent undamped, underdamped, critically damped, and overdamped systems.

A coordinate graph of x(t) vs t showing four curves: undamped (black), underdamped (green), critically damped (blue), and overdamped (red).

Key Observations

  • Undamped: A pure sinusoidal wave that maintains a constant amplitude over time.
  • Underdamped: An oscillating wave where the amplitude decays exponentially over time.
  • Critically Damped: The system returns to equilibrium as quickly as possible without oscillating.
  • Overdamped: The system returns to equilibrium slowly without oscillating.