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MATH 461: Fourier Series and Boundary ValueProblems
Chapter IV: Vibrating Strings and Membranes
Greg Fasshauer
Department of Applied MathematicsIllinois Institute of Technology
Fall 2015
fasshauer@iit.edu MATH 461 Chapter 4 1
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Outline
1 Derivation of Vertically Vibrating Strings
2 Boundary Conditions
3 Example: Vibrating String with Fixed Ends
4 Membranes (omitted now, done later in Chapter 7)
fasshauer@iit.edu MATH 461 Chapter 4 2
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Derivation of Vertically Vibrating Strings
Vibrating Strings
We will now derive a new mathematical model.
Consider a stretched elastic stringof length L with equilibrium positionalong the x-axis.Every point (x ,0), 0 x L, of thestring has a displacement
y = u(x , t)
at any given time t 0.Slow Normal Fast Play/Pause Stop
fasshauer@iit.edu MATH 461 Chapter 4 4
String.aviMedia File (video/avi)
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Derivation of Vertically Vibrating Strings
In order to be able to come up with a reasonable (and manageable)mathematical model we need to make some simplifying assumptions:
We consider only smalldisplacements (relative to thelength of the string).This implies that we canneglect horizontaldisplacements.We assume a perfectly flexiblestring, i.e., only tangentialforces are acting on the string(see the figure).
Here we denote byT the (tangential) tension, the angle T forms with the horizontal (measuredcounter-clockwise).
fasshauer@iit.edu MATH 461 Chapter 4 5
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Derivation of Vertically Vibrating Strings
For a small segment from x to x + x we will use Newtons law
F = m a
coupled with a conservation law that balances the forces in a segmentof the string as
{total force according to Newton}=
{vertical force at left end of segment}+
{vertical force at right end of segment}+
{vertical component of body force (or possible external force)}
orm a = VL + VR + Vbody
to derive a PDE for the displacement u = u(x , t).fasshauer@iit.edu MATH 461 Chapter 4 6
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Derivation of Vertically Vibrating Strings
To this end we needmass:
m(x) = 0(x) density
x
acceleration:
a(x , t) =2ut2
(x , t)
body force:Vbody (x , t) = 0(x)x Q(x , t)
with Q(x , t) the vertical component of the body force per unit mass
A commonly used body force is Vbody = m g.
fasshauer@iit.edu MATH 461 Chapter 4 7
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Derivation of Vertically Vibrating Strings
We also need to carefully study the tensile forces in the string.
Since the displacement and motion are assumed to happen only in thevertical direction we require only the vertical component of the tensileforce:
At the left end:
VL(x , t) = T (x , t) sin (x , t)
At the right end:
VR(x , t) = T (x+x , t) sin (x+x , t)
fasshauer@iit.edu MATH 461 Chapter 4 8
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Derivation of Vertically Vibrating Strings
Putting all of this together, the balance of force equation gives us
0(x)x2ut2
(x , t) = VL(x , t) + VR(x , t) + 0(x)x Q(x , t)
= T (x , t) sin (x , t) + T (x + x , t) sin (x + x , t)+0(x)x Q(x , t)
If we divide by x and let x go to zero we get
0(x)2ut2
(x , t) =T (x + x , t) sin (x + x , t) T (x , t) sin (x , t)
x
x [T (x ,t) sin (x ,t)] as x0
+0(x)Q(x , t)i.e.,
0(x)2ut2
(x , t) =
x[T (x , t) sin (x , t)] + 0(x)Q(x , t).
fasshauer@iit.edu MATH 461 Chapter 4 9
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Derivation of Vertically Vibrating Strings
Since the vertical displacement is assumed to be small, the angle(x , t) is also small (so that cos (x , t) 1).Therefore,
sin (x , t) sin (x , t)cos (x , t)
= tan (x , t)
Here it is important to note that tan (x , t) corresponds to the slope ofthe string, i.e.,
tan (x , t) =ux
(x , t).
Thus
0(x)2ut2
(x , t) =
x[T (x , t) sin (x , t)] + 0(x)Q(x , t)
turns into the PDE for the vibrating string
0(x)2ut2
(x , t) =
x
[T (x , t)
ux
(x , t)]
+ 0(x)Q(x , t).
fasshauer@iit.edu MATH 461 Chapter 4 10
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Derivation of Vertically Vibrating Strings
In order to get the commonly used version of the 1D wave equation wemake two further assumptions:
Small displacements and perfectly elastic strings imply anapproximately constant tension, i.e.,
T (x , t) T0 = const.
So
0(x)2ut2
(x , t) = T02ux2
(x , t) + 0(x)Q(x , t)
Often the external force term can be neglected because theweight of the string is small compared to the tension so that thereis no sag in the string. Then we have
2ut2
(x , t) = c22ux2
(x , t) with c2 =T00(x)
,
the 1D wave equation.
fasshauer@iit.edu MATH 461 Chapter 4 11
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Derivation of Vertically Vibrating Strings
Since the 1D wave equation containsa second-order spatial derivative term anda second-order time derivative term
we need to specifytwo spatial conditions (as usual, we do this in the form ofboundary conditions see next section), andtwo temporal conditions. These will be given as
initial position: u(x ,0) = f (x) and
initial velocity:ut
(x ,0) = g(x).
fasshauer@iit.edu MATH 461 Chapter 4 12
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Boundary Conditions
Fixed Ends
The simplest form of boundary conditions is
u(0, t) = 0u(L, t) = 0
We could also have ends that are affixed to a moving support. Thenthey would be of the form
u(0, t) = f1(t)u(L, t) = f2(t)
fasshauer@iit.edu MATH 461 Chapter 4 14
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Boundary Conditions
Elastic Ends
This will occur if a string is, e.g., attached to a spring-mass system
Thusu(0, t) = y(t),
where y(t) is unknown. In fact, y is determined by an ODE for aspring-mass system with a possibly moving support ys(t).Note that to keep things manageable we assume that thespring-mass system moves only vertically.
fasshauer@iit.edu MATH 461 Chapter 4 15
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Boundary Conditions
The ODE for y is obtained by combining Newtons and Hookes laws:
md2ydt2
(t) = k [y(t) ys(t) `] + other forces,
wherek is the spring constant,ys(t) denotes the moving support of the spring, possibly driven bysome external force,` is the length of the unstretched spring.
The other forces should include the vertical component of the tensileforce of the string:
T (0, t) sin (0, t)small T0 tan (0, t) = T0
ux
(0, t).
Thus, elastic BCs are of the form
md2ydt2
(t) = k [y(t) ys(t) `] + T0ux
(0, t) + g(t).
fasshauer@iit.edu MATH 461 Chapter 4 16
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Boundary Conditions
In the special caseg(t) = 0 (i.e., no additional external forces)with a spring-mass system with small mass, i.e., m 0,and taking y(t) = u(0, t)
we get the BC
k [u(0, t) ys(t) `] = T0ux
(0, t)
k [u(0, t) yE (t)] = T0ux
(0, t),
where yE (t) = ys(t) + ` denotes the equilibrium position of thespring-mass system.
fasshauer@iit.edu MATH 461 Chapter 4 17
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Boundary Conditions
If in addition yE (t) = 0, i.e., the equilibrium position of the spring-masssystem coincides with the equilibrium position of the string and occursat the origin, then
k [u(0, t) yE (t)] = T0ux
(0, t)
becomesT0ux
(0, t) = ku(0, t).
RemarkNote that this BC is analogous to the one we obtained via Newtonslaw of cooling for the 1D heat equation.
fasshauer@iit.edu MATH 461 Chapter 4 18
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Boundary Conditions
RemarkNote that the tensile force and spring constant are usually positive, i.e.,T0 > 0, k > 0. Then we see from
T0ux
(0, t) = ku(0, t) that u(0, t) > 0 ux
(0, t) > 0.
However, at the right end, x = L, we will have
u(L, t) > 0 ux
(L, t) < 0,
i.e., upward motion pulls the string inside. Therefore, thecorresponding BC at the right end has a different sign and looks like
T0ux
(L, t) = ku(L, t).
fasshauer@iit.edu MATH 461 Chapter 4 19
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Boundary Conditions
Free Ends
This will happen when the string is attached to a ring that slidesfrictionless along a vertical support.
Physically, this corresponds to the limiting case of the spring-masssystem for k 0, i.e.,
T0ux
(0, t) = 0.
RemarkNote the analogy to the insulated end condition for the heat equation.
fasshauer@iit.edu MATH 461 Chapter 4 20
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Example: Vibrating String with Fixed Ends
We solve the PDE
2
t2u(x , t) = c2
2
x2u(x , t), for 0 < x < L, t > 0, (1)
with boundary conditions
u(0, t) = u(L, t) = 0 for t > 0 (2)
and initial conditions
u(x ,0) = f (x) for 0 < x < L (initial position) (3)ut
(x ,0) = g(x) for 0 < x < L (initial velocity) (4)
From our earlier derivations the constant c2 = T00 is given as the ratioof tension to density.
fasshauer@iit.edu MATH 461 Chapter 4 22
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Example: Vibrating String with Fixed Ends
RemarkSince the PDE and its BCs are linear and homogeneous we attempt tosolve the problem using separation of variables.
The usual Ansatz u(x , t) = (x)T (t) gives us the partial derivatives
2
t2u(x , t) = (x)T (t)
2
x2u(x , t) = (x)T (t)
so that we have(x)T (t) = c2(x)T (t)
or1c2
T (t)T (t)
=(x)(x)
= .
fasshauer@iit.edu MATH 461 Chapter 4 23
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Example: Vibrating String with Fixed Ends
The resulting two ODEs are
T (t) = c2T (t) (5)
and(x) = (x) (6)
with BCs(0) = (L) = 0. (7)
RemarkWe chose for the separation constant above so that (6)-(7) is oneof our standard boundary-value problems with well-known eigenvaluesand eigenfunctions
n =(n
L
)2, n(x) = sin
nxL, n = 1,2,3, . . .
fasshauer@iit.edu MATH 461 Chapter 4 24
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Example: Vibrating String with Fixed Ends
RemarkMoreover, the ODE (5) for the time-dependent component T has thephysically meaningful, oscillating solution for positive , i.e.,
T (t) = c1 cos ct + c2 sin c
t .
For this example the other two types of solutionT (t) = c1t + c2 for = 0
T (t) = c1ect + c2ec
t for < 0
are not relevant.
fasshauer@iit.edu MATH 461 Chapter 4 25
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Example: Vibrating String with Fixed Ends
Using the principle of superposition to combine the solutions of theODEs (5) and (6) we get
u(x , t) =
n=1
[An cos
cntL
+ Bn sincnt
L
]sin
nxL
To determine the expansion coefficients An and Bn we use the initialconditions (3) and (4).Since cos 0 = 1 and sin 0 = 0 we have
u(x ,0) =
n=1
An sinnx
L!
= f (x)
so that
An =2L
L0
f (x) sinnx
Ldx .
fasshauer@iit.edu MATH 461 Chapter 4 26
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Example: Vibrating String with Fixed Ends
In order to apply the initial condition (4) we need to ensure thatu is continuous andut is piecewise smooth.
Then we can apply term-by-term differentiation to get
ut
(x , t) =
n=1
[An
cnL
sincnt
L+ Bn
cnL
coscnt
L
]sin
nxL
andut
(x ,0) =
n=1
Bncn
Lsin
nxL
!= g(x)
so that
Bn =2
cn
L0
g(x) sinnx
Ldx .
The solution for this problem is illustrated in the Mathematica notebookWave.nb.
fasshauer@iit.edu MATH 461 Chapter 4 27
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Example: Vibrating String with Fixed Ends
Normal Modes and Overtones: Applications to Music
For each n, the n-th term of the Fourier series solution
un(x , t) =[An cos
cntL
+ Bn sincnt
L
]sin
nxL
is called the n-th normal mode of the solution.
The term un describes a harmonic motion with frequency
fn =cn
L/2 =
cn2L
or with circular frequency = cnL .
The frequencies fn are called the natural frequencies of the solution u.
fasshauer@iit.edu MATH 461 Chapter 4 28
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Example: Vibrating String with Fixed Ends
Figure: Third mode, or second overtone, n = 3.
fasshauer@iit.edu MATH 461 Chapter 4 29
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Example: Vibrating String with Fixed Ends
Note that each one of the modes will be zero for all t if
sinnx
L= 0, i.e.,
nxL
= k, k = 1,2,3, . . .
The points
xk = kLn, k = 1,2, . . . ,n
are called nodes.
By placing their finger at a node point, players of string instrumentssuch as guitars, violins, etc., can produce so-called flageolet tones.
fasshauer@iit.edu MATH 461 Chapter 4 30
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Example: Vibrating String with Fixed Ends
A string instrument can be tuned, i.e., the pitch can be made higher bydecreasing the length of the string, L, or
increasing c. Since c =
T00
this means that one can eitherincrease the tension ordecrease the density of the string.
RemarkOvertones are illustrated acoustically in the MATLAB scriptovertones.m.
We can hear there that the first overtone (i.e., with double thefrequency) is an octave higher than the fundamental mode.Similarly, the second overtone is a fifth higher than the firstovertone,and the third overtone is a fourth higher than the second overtone.
fasshauer@iit.edu MATH 461 Chapter 4 31
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Example: Vibrating String with Fixed Ends
Traveling Waves (Exercise 4.4.7)Lets again consider the 1D wave equation
2
t2u(x , t) = c2
2
x2u(x , t)
u(0, t) = u(L, t) = 0
u(x ,0) = f (x) andut
(x ,0) = g(x) = 0
From the general solution derived above we know that
Bn =2
cn
L0
g(x) sinnx
Ldx = 0.
Therefore
u(x , t) =
n=1
An coscnt
Lsin
nxL
with
An =2L
L0
f (x) sinnx
Ldx .
fasshauer@iit.edu MATH 461 Chapter 4 32
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Example: Vibrating String with Fixed Ends
The trigonometric identity
sin cos =12
[sin( + ) + sin( )]
with = nxL and =cnt
L gives us
coscnt
Lsin
nxL
=12
[sin(
nxL
+cnt
L
)+ sin
(nx
L cnt
L
)]=
12
[sin
nL
(x + ct) + sinnL
(x ct)]
So the solution
u(x , t) =
n=1
An coscnt
Lsin
nxL
becomes
u(x , t) =12
n=1
An sinnL
(x + ct) +12
n=1
An sinnL
(x ct)
fasshauer@iit.edu MATH 461 Chapter 4 33
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Example: Vibrating String with Fixed Ends
RemarkNote that the two parts of
u(x , t) =12
n=1
An sinnL
(x + ct) +12
n=1
An sinnL
(x ct)
are Fourier sine series of f evaluated at x + ct and x ct, respectively.
Thereforeu(x , t) =
12
[f (x + ct) + f (x ct)
],
where f is the odd 2L-periodic extension of f .
RemarkThis shows that we can get the solution of the 1D wave equationwithout actually summing the infinite series!
fasshauer@iit.edu MATH 461 Chapter 4 34
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Example: Vibrating String with Fixed Ends
The solution of the 1D wave equation in the form
u(x , t) =12
[f (x + ct) + f (x ct)
],
is known as dAlemberts solution, after Jean dAlembert who firstformulated the 1D wave equation and proposed this form of thesolution in 1746 22 years before Fouriers birth.It can be interpreted as the average of two traveling waves:
one traveling to the left,the other to the right
both with speed c.This is illustrated in the Mathematica notebook Wave.nb.
RemarkThe traveling wave solution is closely related to the solution of PDEsby the method of characteristics (see Chapter 12 in [Haberman] andMATH 489).
fasshauer@iit.edu MATH 461 Chapter 4 35
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Appendix References
References I
R. Haberman.Applied Partial Differential Equations.Pearson (5th ed.), Upper Saddle River, NJ, 2012.
fasshauer@iit.edu MATH 461 Chapter 4 37
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Derivation of Vertically Vibrating StringsBoundary ConditionsExample: Vibrating String with Fixed EndsMembranes (omitted now, done later in Chapter 7)Appendix
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