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PAGE 1

Wave Equation

models a vibrating string

string of length \(L\), held tight, ends fixed

A straight horizontal line segment representing a string in its natural position. The left end is labeled x=0 and the right end is labeled x=L.
natural position
A curved line representing a displaced string above a dashed horizontal line representing the natural position y=0. The left end is x=0 and the right end is x=L. An arrow points down from the curve to the dashed line, labeled y(x,t).
\(y(x,t)\) : displacement from natural position

the string naturally "wants" to be flat and will provide a restoring

acceleration / force

if displaced like this :

A simple concave down curve representing a displaced string.

\(y_{xx} < 0\) , string "fights" back

w/ a downward acceleration \(\rightarrow y_{tt} < 0\)

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if displaced like this:

A small sketch of an upward-opening concave arc.

\( y_{xx} > 0 \), string has

an upward acceleration \( \rightarrow y_{tt} > 0 \)

acceleration and concavity have the same sign.

\[ \frac{\partial^2 y}{\partial t^2} = a^2 \frac{\partial^2 y}{\partial x^2} \]

1-D Wave Eq.

\[ y_{tt} = a^2 y_{xx} \]

\( a^2 \) determines how fast string vibrates (wave speed)

for string: \( a^2 = \frac{\text{tension}}{\text{density}} \)

Boundary conditions:

  • \( y(0,t) = y(L,t) = 0 \) ends fixed at 0 (Dirichlet)
  • or \( y_x(0,t) = y_x(L,t) = 0 \) ends can move vertically but not horizontally (Neumann)
A sketch showing a wave between two vertical dashed lines with arrows indicating vertical movement at the boundaries.
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Initial Conditions

because 2nd-order deriv. with time

  • \( y(x, 0) = f(x) \) initial displacement (plucking)
  • \( y_t(x, 0) = g(x) \) initial velocity (strumming)

today we will solve the problem w/ ends fixed at 0 and no initial velocity ("Problem A")

\[ y_{tt} = a^2 y_{xx} \]
\( 0 < x < L \quad t > 0 \)
\( y(0, t) = y(L, t) = 0 \)
ends fixed at 0
\( y_t(x, 0) = g(x) = 0 \)
no initial velocity
\( y(x, 0) = f(x) \)
initial displacement

we will again use the method of separation of variables

\[ y(x, t) = X(x)T(t) \]

\( y_{xx} = X''T \)

\( y_{tt} = XT'' \)

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Wave Equation Separation of Variables

wave eq:
\[ \bar{X} T'' = a^2 \bar{X}'' T \]
\[ \frac{\bar{X}''}{\bar{X}} = \frac{T''}{a^2 T} = \text{separation constant} = -\lambda \]
(just like w/ heat eq.)

we get two ODEs:

\[ \bar{X}'' + \lambda \bar{X} = 0 \] \[ T'' + a^2 \lambda T = 0 \]
BC:
\[ y(0,t) = 0 \rightarrow \bar{X}(0)T(t) = 0 \rightarrow \bar{X}(0) = 0 \]
\[ y(L,t) = 0 \rightarrow \bar{X}(L)T(t) = 0 \rightarrow \bar{X}(L) = 0 \]

\(\bar{X}\) solution is the same as non-insulated heat eq

\(\vdots\)
\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \]
eigenvalues
\(n = 1, 2, 3, \dots\)
\[ \bar{X}_n = \sin\left(\frac{n \pi x}{L}\right) \]
eigenfunctions
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\[ T'' + a^2 \lambda T = 0 \]

\[ T'' + \frac{a^2 n^2 \pi^2}{L^2} T = 0 \]

\[ T(t) = A \cos\left(\frac{n \pi a}{L} t\right) + B \sin\left(\frac{n \pi a}{L} t\right) \]

IC:

\( y_t(x, 0) = 0 \)

no initial velocity

\( X(x) T'(0) = 0 \rightarrow T'(0) = 0 \)

\(\vdots\)

\( B = 0 \)

\[ T_n = \cos\left(\frac{n \pi a}{L} t\right) \]

\( n = 1, 2, 3, \dots \)

for wave, the time solution is also periodic

for each \( n \), there is one solution \( y_n = X_n T_n \)

each of these is called

a "mode" or "harmonic"

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general solution:

\[ y(x, t) = \sum_{n=1}^{\infty} A_n \cos\left(\frac{n \pi a}{L} t\right) \sin\left(\frac{n \pi}{L} x\right) \]

last IC:

\( y(x, 0) = f(x) \)

initial displacement

\[ f(x) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n \pi x}{L}\right) \]

sine series

\[ A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) dx \]

example

\( L = 20 \),

\( a = 1 \),

\( g(x) = 0 \) no initial velocity

initial displacement:

\[ f(x) = \begin{cases} \frac{1}{5}x & 0 < x < 5 \\ 1 & 5 < x < 15 \\ \frac{20-x}{5} & 15 < x < 20 \end{cases} \]

A graph of the piecewise function f(x) representing initial displacement. The horizontal axis is labeled x and the vertical axis is labeled f(x). The function starts at the origin, increases linearly to a height of 1 at x=5, stays flat at y=1 until x=15, and then decreases linearly back to the x-axis at x=20.
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Harmonic Analysis of Wave Equation Solution

\[ y(x,t) = \sum_{n=1}^{\infty} \frac{8}{n^2\pi^2} \left[ \sin\left(\frac{n\pi}{4}\right) + \sin\left(\frac{3n\pi}{4}\right) \right] \cos\left(\frac{n\pi}{20} t\right) \sin\left(\frac{n\pi}{20} x\right) \]

\( n=1 \) Fundamental mode / harmonic

\[ y_1(x,t) = \frac{8\sqrt{2}}{\pi^2} \cos\left(\frac{\pi}{20} t\right) \sin\left(\frac{\pi}{20} x\right) \]

\(\rightarrow\) frequency of vibration (sound)

\( \frac{\pi}{20} \text{ rad/s} \approx \frac{\pi/20}{2\pi} = \frac{1}{40} \text{ Hz} \)

\( n=2 \) second harmonic

freq: \( 2 \cdot \frac{\pi}{20} = \frac{1}{20} \text{ Hz} \)

(doubles first harmonic)

\(\rightarrow\) an octave higher

\( n=3 \) 3rd harmonic

freq: \( 3 \cdot \frac{\pi}{20} = \frac{3}{40} \text{ Hz} \)

(one octave and a perfect fifth above fundamental)

what we hear is combination of ALL n's

PAGE 8

Surface Plot: \( y(x,t) \)

A 3D surface plot showing displacement y as a function of position x and time t. The surface has a wave-like form with peaks and valleys, colored with a gradient from blue (low) to yellow (high). Position x ranges from 0 to 20, time t ranges from 0 to 40, and displacement y ranges from -1.00 to 1.00.
PAGE 9

String Displacement at Snapshots in Time (Ì‚y vs x)

The following graph illustrates the displacement of a string at various discrete time intervals. The horizontal axis represents the position along the string (Ì‚x), and the vertical axis represents the displacement (Ì‚y).

A line graph showing string displacement y versus position x at six different time snapshots: t=0, t=4, t=8, t=12, t=16, and t=20. The curves show a wave-like pattern that evolves from a trapezoidal shape at t=0 to an inverted trapezoidal shape at t=20, passing through a flat line at t=10 (not explicitly labeled but implied by the symmetry).

Legend and Observations

  • t = 0: Initial trapezoidal displacement reaching a maximum of 1.00 between x=5 and x=15.
  • t = 4: The displacement begins to decrease and narrow.
  • t = 8: Further reduction in amplitude, peaking at approximately 0.4.
  • t = 12: The displacement becomes negative, mirroring the earlier positive shapes.
  • t = 16: The negative displacement reaches a minimum of -1.00 between x=9 and x=11.
  • t = 20: The displacement forms an inverted trapezoid, reaching -1.00 between x=5 and x=15.
PAGE 10

Oscillation of Specific Points (Ì‚y vs t)

This graph tracks the displacement (Ì‚y) over time (t) for three specific points along the string: x=5, x=10, and x=15. This provides a temporal view of the oscillation at fixed spatial locations.

A line graph showing displacement y versus time t for three specific positions: x=5, x=10, and x=15. The point x=10 (yellow line) stays at y=1.0 until t=5, then drops linearly to y=-1.0 at t=15, stays there until t=25, and returns to y=1.0 at t=35. The points x=5 and x=15 (blue and green lines) follow a triangular path, dropping from y=1.0 at t=0 to y=-1.0 at t=20 and returning to y=1.0 at t=40.

Point Analysis

  • x = 5 and x = 15: These points exhibit identical behavior, following a V-shaped oscillation with a period of 40 units.
  • x = 10: This point, located at the center of the initial trapezoid, maintains its maximum displacement for a short duration before oscillating, resulting in a trapezoidal wave in time.