The document discusses the derivation of the wave equation from first principles. It begins by considering transverse vibrations on an elastic string under tension. By analyzing changes in tension and angle as a small segment of string is displaced, an expression for the wave motion is obtained. Approximating for small motions, the standard one-dimensional wave equation is derived. The derivation is then extended to multiple dimensions. Applications to damped systems and transmission lines are also discussed. Worked examples are provided to solve the wave equation for specific initial conditions.

1. By:
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2. Presented By:
Joseph Ash
Jordan Baldwin
Justin Hirt
Andrea Lance
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3. Jean Baptiste Biot
 (1774-1862)
 French Physicist
 Worked on analysis of
heat conduction
 Unsuccessful at dealing
with the problem of
incorporating external
convection effects in heat
conduction analysis
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4. Jean Baptiste Joseph Fourier
 (1768 – 1830)
 Read Biot’s work
 1807 determined how to solve the
problem
 Fourier’s Law
 Time rate of heat flow (Q) through a slab
is proportional to the gradient of
temperature difference
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5. Ernst Schmidt
 German scientist
 Pioneer in Engineering Thermodynamics
 Published paper “Graphical Difference
Method for Unsteady Heat Conduction”
 First to measure velocity and
temperature field in free convection
boundary layer and large heat transfer
coefficients
 Schmidt Number
 Analogy between heat and mass
transfer that causes a dimensionless
quantity
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6. A first approximation of the equations that govern the
conduction of heat in a solid rod.
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7.  A uniform rod is insulated on both lateral ends.
 Heat can now only flow in the axial direction.
 It is proven that heat per unit time will pass
from the warmer section to the cooler one.
 The amount of heat is proportional to the area,
A, and to the temperature difference T2-T1, and
is inversely proportional to the separation
distance, d.
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8.  The final consideration can be expressed as the
following:
is a proportionality factor called the thermal
conductivity and is determined by material properties
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9.  The bar has a length L so x=0 and x=L
 Perfectly insulated
 Temperature, u, depends only on position, x,
and time, t
 Usually valid when the lateral dimensions are small
compared to the total length.
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10.  The differential equation governing the
temperature of the bar is a physical
balance between two rates:
 Flux/Flow term
 Absorption term
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11.  The instantaneous rate of heat transfer from left to right
across the cross sections x=x0 where x0 is arbitrary can be
defined as:
 The negative is needed in order to show a positive
rate from left to right (hot to cold)
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12.  Similarly, the instantaneous rate of heat transfer from
right to left across the cross section x=x0+Δx where Δx
is small can be defined as:
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13.  The amount of heat entering the bar in a time span of
Δt is found by subtracting the previous two equations
and then multiplying the result by Δt:
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14.  The average change in temperature, Δu, can be
written in terms of the heat introduced, Q Δt and the
mass Δm of the element as:
where s = specific heat of the material
ρ = density
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15.  The actual temperature change of the bar is simply
the actual change in temperature at some
intermediate point, so the above equation can also
be written as:
This is the heat absorption equation.
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16.  Equating the QΔt in the flux and absorption terms,
we find the heat absorption equation to be:
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17.  If we divide the above equation by ΔxΔt and allow both
Δx and Δt to both go to 0, we will obtain the heat
conduction or diffusion equation:
where
and has the dimensions of length^2/time and called
the thermal diffusivity
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18.  Certain boundary conditions may apply to the
specific heat conduction problem, for example:
 If one end is maintained at some constant
temperature value, then the boundary condition for
that end is u = T.
 If one end is perfectly insulated, then the boundary
condition stipulates ux = 0.
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19.  Consider the end where x=0 and the rate of flow of heat
is proportional to the temperature at the end of the bar.
 Recall that the rate of flow will be given, from left to right, as
 With this said, the rate of heat flow out of the bar from right to
left will be
 Therefore, the boundary condition at x=0 is
where h1 is a proportionality constant
if h1=0, then it corresponds to an insulated end
if h1 goes to infinity, then the end is held at 0 temp.
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20.  Similarly, if heat flow occurs at the end x = L, then the
boundary condition is as follows:
where, again, h2 is a nonzero proportionality factor
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21.  Finally, the temperature distribution at one
fixed instant – usually taken at t = 0, takes
the form:
occurring throughout the bar
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22.  Sometimes, the thermal conductivity, density, specific
heat, or area may change as the axial position
changes. The rate of heat transfer under such
conditions at x=x0 is now:
 The heat equation then becomes a partial differential
equation in the form:
or
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23.  Other ways for heat to enter or leave a bar must also
be taken into consideration.
 Assume G(x,t,u) is a rate per unit per time.
 Source
 G(x,t,u) is added to the bar
 G(x,t,u) is positive, non-zero, linear, and u does not depend on t
 G(x,t,u) must be added to the left side of the heat equation
yielding the following differential equation
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24.  Similarly,
 Sink
 G(x,t,u) is subtracted from the bar
 G(x,t,u) is positive, non-zero, linear, and u does not
depend on t
 G(x,t,u) then under this sink condition takes the form:
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25.  Putting the source and sink equations
together in the heat equation yields
which is commonly called the generalized
heat conduction equation
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26.  Now consider a bar in which the temperature is a
function of more than just the axial x-direction.
Then the heat conduction equation can then be
written:
 2-D:
 3-D:
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27. Let an aluminum rod of length 20 cm be initially at the
uniform temperature 25°C. Suppose that at time t=0,
the end x=0 is cooled to 0°C while the end x=20 is
heated to 60°C, and both are thereafter maintained at
those temperatures.
Find the temperature distribution in
the rod at any time t
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28. Find the temperature distribution, u(x,t)
a2uxx=ut, 0<x<20, t<0
u(0,t)=0 u(20,t)=60, t<0
u(x,0)=25, 0<x<20
From the initial equation we find that:
L=20, T1=0, T2=60, f(x)=25
We look up the Thermal Diffusivity of aluminum→a2=0.86
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29. Using Equations 16 and 17 found on page 614, we find
that
where
u x t T T x p p a
T c e n x
( ) ( ) å¥
n t
2 1 1 , sin 2
=
-
ö çè
÷ø
= - + + æ
1
2 2 2
n
L
n L
L
T n x
f x T T x
= é - - - L
n dx
çè
æ
ö ù
( ) ( ) ò ÷ø
úû
êë
L
L
L
c
0 2 1 1 2 sin p
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30. Evaluating cn, we find that
c x n x dx
= ò é - ( - )
-
ö çè
ù
( ( ) ( ) )
2
é - + =
c n n n n
p p p p
10 7 cos 12sin 5
( )
( ( ) )
n
c n
p
p
p
p
n
n
n
L
n
70cos 50
20
0 sin
20
25 60 0
20
2
0
= +
ù
úû
êë
÷ø
æ
úû
êë
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31. Now we can solve for u(x,t)
( ) ( ) ( ( ) )
( ) ( ( ) ) ( )
u x t x n
ö çè= - + + æ +
u x t x n
å
å
¥
=
-
¥
=
-
2 2 2
0.86
ö çè
÷ø
p
ö çè
p
0.86
æ ÷ø
= + æ +
ö çè
÷ø
æ ÷ø
1
400
1
20
20
, 3 70cos 50 sin
20
0 70cos 50 sin
20
, 60 0
2
2
n
n t
n
n t
e n x
n
e n x
n
p
p
p
p
p
p
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33. Derivation of the Wave Equation
Applicable for:
•One space dimension, transverse vibrations on elastic string
•Endpoints at x = 0 and x = L along the x-axis
•Set in motion at t = 0 and then left undisturbed
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34. Schematic of String in Tension
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35. Since there is no acceleration in the horizontal direction
T(x + Dx, t) cos(q + Dq ) - T(x, t) cosq = 0
However the vertical components must satisfy
T(x x,t) sin( ) T(x,t)sin xu (x,t) tt +D q + Dq - q = rD
x
where is the coordinate to the center of mass and the
weight is neglected
Replacing T with V the and rearranging the equation becomes
V x + D x t -
V x t
( , ) ( , ) u (x, t)
x
tt = r
D
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36. Letting , the equation becomes
Dx®0
V (x, t) u (x, t) x tt = r
To express this in terms of only terms of u we note that
V (x,t) H(t) tan H(t)u (x,t) x = q =
The resulting equation in terms of u is
x x tt (Hu ) = ru
and since H(t) is not dependant on x the resulting equation is
xx tt Hu = ru
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37. For small motions of the string, it is approximated that
H = T cosq » T
using the substitution that
a2 = T / r
the wave equation takes its customary form of
xx tt a2u = u
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38. The telegraph equation
u cu ku a2u F(x,t) tt t xx + + = +
where c and k are nonnegative constants
cut arises from a viscous damping force
ku arises from an elastic restoring force
F(x,t) arises from an external force
The differences between this telegraph equation and the customary
wave equation are due to the consideration of internal elastic
forces. This equation also governs flow of voltage or current in a
transmission line, where the coefficients are related to the electrical
parameters in the line.
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39. For a vibrating system with more than on significant space coordinate it
may be necessary to consider the wave equation in more than one
dimension.
For two dimensions the wave equation becomes
xx yy tt a2 (u + u ) = u
For three dimensions the wave equation becomes
xx yy zz tt a2 (u + u + u ) = u
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40. Consider an elastic string of length L whose ends are
held fixed. The string is set in motion from its
equilibrium position with an initial velocity g(x). Let
L=10 and a=1. Find the string displacement for any
time t.
( )
ì
ï ï
x
4 ,
1,
í
( L -
x
ï ) ï
î
=
4 ,
L
L
g x
x L
£ £
L < x <
3
L
L £ x £
L
3
4
4
4
4
0
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41. From equations 35 and 36 on page 631, we find that
u x t k npx p
where ( ) å¥ =
n at
ö çè
÷ø
ö çè
æ ÷ø
= æ
, sin sin
1
n
n L
L
g x n x
p 2 sin p
= æ L
n dx
ö çè
ò ÷ø
( ) L
L
n a
k
L
0
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42. Solving for kn, we find:
( )
é
3
n x
dx L x
dx n x
n x
x
n a
p p p p
2 4 sin sin 4 sin
= ò æ ò ò
( )
( )
æ + ÷ø
ö çè
ö
æ
ö çè
æ
æ ÷ø
ö çè
n
k L
ö çè= æ
( ) ÷ ÷ø
ç çè
÷ø
æ + ÷ø
ö
÷ ÷ø
ç çè
- ÷ø
æ + ÷ø
= æ
ù
úû
êë
ö çè
÷ø
æ - + ÷ø
ö çè
ö çè
ö çè
4
sin
4
8 sin 3
sin
4
sin
4
2 4 sin 3
3
2
4
0
4
3
4 4
p p
p
p p p
p p
n n
a n
n n n
n
L
n a
k
dx
L
L
L
L
L
L
k
L
n
L L
L
L
n L
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43. Now we can solve for u(x,t)
u x t L
( )
( )
¥
æ
( )
å
( ) å
å
3 3
=
1
n
¥
3 3
=
¥
=
n at
ö çè
n at
ö çè
ö çè
÷ø
n n n x
ö çè
n n n x
ö çè
ö çè
æ ÷ø
æ
ö
ö
ö
÷ ÷ø
æ
æ
ç çè
ö çè
ö
ö
÷ ÷ø
æ
ç çè
ö çè
ö çè
÷ø
ö çè
ö çè
ö çè
æ + ÷ø
æ
= æ
÷ø
æ ÷ø
æ
÷ ÷ø
ç çè
÷ ÷ø
ç çèæ
÷ø
æ + ÷ø
= æ
÷ø
æ ÷ø
æ
ö
÷ ÷ø
ç çè
÷ ÷ø
ç çè
÷ø
æ + ÷ø
= æ
1
1
3
10
sin
10
sin
4
sin
4
, 80 1 sin 3
sin sin
4
sin
4
, 8 1 sin 3
sin sin
4
sin
4
, 8 sin 3
n
n
n n n x n t
n
u x t
L
L
n
u x t L
L
L
a n
p p p p
p
p p p p
p
p p p p
p
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