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PAGE 1

Wave Eq. (part 2)

\[ y_{tt} = a^2 y_{xx} \quad 0 < x < L \quad t > 0 \]

  • \( y(0,t) = y(L,t) = 0 \) ends fixed
  • \( y(x,0) = f(x) \) initial displacement
  • \( y_t(x,0) = g(x) \) initial velocity
A diagram showing a string of length L fixed at both ends x=0 and x=L. A dashed line represents the equilibrium position y=0, and a curved solid line above it represents the displacement y(x,t).

last time: \( g(x) = 0 \) (no initial velocity)

\[ y(x,t) = \sum_{n=1}^{\infty} A_n \cos\left(\frac{n\pi at}{L}\right) \sin\left(\frac{n\pi x}{L}\right) \] \[ A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \]

today:

  • \( f(x) = 0 \) (no initial displacement)
  • \( g(x) \neq 0 \) some initial velocity ("Problem B")
PAGE 2

same basic idea: \( y(x,t) = X(x) T(t) \)

\( \vdots \) \[ \frac{X''}{X} = \frac{T''}{a^2 T} = -\lambda \]

two ODEs:

  • \( X'' + \lambda X = 0 \)
  • \( T'' + a^2 \lambda T = 0 \)

same BCs:

  • \( y(0,t) = 0 \rightarrow X(0) = 0 \)
  • \( y(L,t) = 0 \rightarrow X(L) = 0 \)

same spatial solution:

\[ \lambda_n = \frac{n^2 \pi^2}{L^2} \quad n=1,2,3,... \]

\[ X_n = \sin\left(\frac{n\pi x}{L}\right) \]

\[ T'' + a^2 \lambda T = 0 \] \[ T'' + \frac{a^2 n^2 \pi^2}{L^2} T = 0 \]
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\[ T(t) = A \cos\left(\frac{n\pi a}{L}t\right) + B \sin\left(\frac{n\pi a}{L}t\right) \]
IC: \( y(x,0) = 0 \rightarrow X(x)T(0) = 0 \rightarrow T(0) = 0 \) (last time: \( T'(0) = 0 \))

here, \( A = 0 \)

so, \( T_n = \sin\left(\frac{n\pi a}{L}t\right) \)

(Problem A has \( \cos\left(\frac{n\pi a}{L}t\right) \))

for each \( n \), \( y_n = \sin\left(\frac{n\pi a}{L}t\right) \sin\left(\frac{n\pi}{L}x\right) \)

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

last IC:

\( y_t(x,0) = g(x) \) initial velocity
\[ y_t(x,t) = \sum_{n=1}^{\infty} \frac{n\pi a}{L} B_n \cos\left(\frac{n\pi a}{L}t\right) \sin\left(\frac{n\pi}{L}x\right) \]
\[ g(x) = \sum_{n=1}^{\infty} \left( \frac{n\pi a}{L} B_n \right) \sin\left(\frac{n\pi}{L}x\right) \]
sine series w/ coefficients \( \frac{n\pi a}{L} B_n \)
\[ \frac{n\pi a}{L} B_n = \frac{2}{L} \int_{0}^{L} g(x) \sin\left(\frac{n\pi}{L}x\right) dx \]
\[ B_n = \frac{2}{n\pi a} \int_{0}^{L} g(x) \sin\left(\frac{n\pi}{L}x\right) dx \]
PAGE 4

example

A string w/ \( L=1 \) and \( a=1 \) is initially at rest and is struck w/ a hammer of width 0.2 at the center w/ upward velocity of 10.

A diagram showing a horizontal string segment between endpoints labeled x=0 and x=1. A rectangular hammer is positioned below the center of the string, pointing upwards.
initial velocity \[ g(x) = \begin{cases} 10 & 0.4 < x < 0.6 \\ 0 & \text{else} \end{cases} \]
A coordinate graph of g(x) versus x. The function is zero everywhere except for a rectangular pulse of height 10 between x=0.4 and x=0.6.
\[ y(x,t) = \sum_{n=1}^{\infty} \frac{20}{n^2\pi^2} \left[ \cos(0.4n\pi) - \cos(0.6n\pi) \right] \sin(n\pi t) \sin(n\pi x) \]
\( \uparrow \)

frequency of each mode (n)

PAGE 5

String Displacement \(y(x, t)\)

The following 3D surface plot illustrates the displacement of a string, denoted as \(y(x, t)\), as a function of both position \(x\) and time \(t\). The visualization shows how a wave pulse propagates and evolves over the spatial domain \(x \in [0, 1]\) through the time interval \(t \in [0, 2]\).

A 3D surface plot showing string displacement y as a function of position x and time t. The surface starts as a narrow peak at t=0 and spreads out into a wider trapezoidal pulse as time increases, with the displacement y ranging from 0 to 1.0.
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Snapshots of String Displacement \(y(x)\) at Different Times

This 2D plot provides a series of snapshots showing the string's displacement \(y\) at specific time intervals. It highlights the transformation of the initial pulse into a broader trapezoidal shape as it moves across the position \(x\).

A 2D line graph showing multiple snapshots of string displacement y versus position x at different times: t=0.1, t=0.2, t=0.3, t=0.5, and t=0.7. The initial narrow triangular pulse at t=0.1 widens into a trapezoidal shape that expands symmetrically outward as time progresses.

Key Observations

  • At \(t = 0.1\), the displacement is a narrow peak centered at \(x = 0.5\).
  • As \(t\) increases to \(0.7\), the pulse widens, maintaining a maximum displacement of \(y = 1.0\) over an increasing range of \(x\).
  • The boundaries of the pulse move outward toward \(x = 0\) and \(x = 1\).
PAGE 7

Motion of Specific String Points \(y(t)\)

The following graph illustrates the displacement \(y\) over time \(t\) for three specific points along a vibrating string, located at positions \(x = 0.2\), \(x = 0.5\), and \(x = 0.8\). The motion exhibits a periodic, trapezoidal wave pattern, characteristic of certain initial conditions in wave mechanics.

A line graph showing displacement y versus time t for three positions: x=0.2, x=0.5, and x=0.8. The curves are periodic trapezoidal waves. The x=0.5 curve (orange) reaches its peak first, followed by x=0.2 (blue) and x=0.8 (teal) which overlap. The displacement ranges from -1.0 to 1.0 over a time interval from 0.0 to 4.0.
PAGE 8

String Displacement at \(t = 0.2\): Total Solution vs. Harmonics

This visualization compares the total solution for string displacement (calculated using \(n=100\) terms) against individual odd harmonics (\(n=1, 3, 5, 7\)) at a fixed time \(t = 0.2\). The total solution shows a flat-topped pulse, while the individual harmonics represent the sinusoidal components that sum to form this shape.

A graph of displacement y versus position x at time t=0.2. A thick black line represents the total solution (n=100), showing a trapezoidal pulse. Several dashed colored lines represent individual harmonics: n=1 (blue), n=3 (red), n=5 (green), and n=7 (yellow). The harmonics are sinusoidal waves of increasing frequency and decreasing amplitude.
PAGE 9

Harmonics in Vibrating Strings

There are no even harmonics in that solution (even \(n\)'s).

Why?

A series of four sketches illustrating the first four harmonic modes (n=1 to n=4) of a vibrating string fixed at both ends. The even modes (n=2 and n=4) clearly show a stationary node at the center of the string.
\(n = 1\)
\(n = 2\) center at rest for all \(t\)
\(n = 3\)
\(n = 4\) center at rest for all \(t\)

Since we are hitting the string at center, the center of string must move, but even harmonics have center at rest so all even harmonics are “destroyed” (same if plucked).

On real piano, the hammer usually strikes at \(L/7\) or \(L/9\)

\(\downarrow\)

eliminate multiples of 7 (dissonant)

PAGE 10

A struck string starts at zero displacement \(\rightarrow\) no initial sound (small delay before sound is made).

Problem A:
displacement only
Problem B:
velocity only

\(y_{tt} = a^2 y_{xx}\) is linear

So general problem: Problem A + Problem B (just add them)

Wind Instruments

For wind instrument, no string, sound is from pressure waves.

e.g. \[ \frac{\partial^2 P}{\partial t^2} = a^2 \frac{\partial^2 P}{\partial x^2} \]
same equation!