math 461: fourier series and boundary value problems - chapter iv: vibrating strings...

MATH 461: Fourier Series and Boundary ValueProblems

Chapter IV: Vibrating Strings and Membranes

Greg Fasshauer

Department of Applied MathematicsIllinois Institute of Technology

Fall 2015

[email protected] MATH 461 Chapter 4 1

http://math.iit.eduhttp://math.iit.edu/~fass

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)



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


String.aviMedia File (video/avi)



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).




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)[email protected] MATH 461 Chapter 4 6



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.




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)




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).




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).




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.




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).



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)



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.



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).



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.



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.



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).



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.



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.




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)

= .




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, . . .




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.




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 .




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.




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.




Figure: Third mode, or second overtone, n = 3.




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.




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.




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 .




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)




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!




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).


http://math.iit.eduhttp://en.wikipedia.org/wiki/Jean_le_Rond_d'Alemberthttp://math.iit.edu/~fass

Appendix References

References I

R. Haberman.Applied Partial Differential Equations.Pearson (5th ed.), Upper Saddle River, NJ, 2012.



Derivation of Vertically Vibrating StringsBoundary ConditionsExample: Vibrating String with Fixed EndsMembranes (omitted now, done later in Chapter 7)Appendix

math 461: fourier series and boundary value problems - chapter iv: vibrating strings...

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