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Continuous Strings, Standing Waves, and Normal Modes

(Topic)

Continuous Strings, Standing Waves, and Normal Modes

A long chain of coupled oscillators approaches a continuous medium when the lattice spacing becomes small compared with the wavelength.

The simplest continuous wave-bearing mechanical system is a stretched string.

Let

For small slopes, the transverse motion obeys

|--------------|
|∂2y      ∂2y  |
|----=  c2----,|
-∂t2------∂x2--
(1)

where

|----∘-----|
|          |
|c =    T. |
|       μ  |
-----------
(2)

This is the one-dimensional wave equation.

A string of length

L
(3)

fixed at both ends cannot support arbitrary single-frequency shapes.

The boundary conditions select a discrete family of normal modes:

|------------(----)--|
|             n-πx   |
Xn-(x-) =-sin--L----,-
(4)

with

|------------------------------|
|      nπc                     |
|ωn =  -L-,     n = 1,2, 3,....|
-------------------------------
(5)

Thus a continuous system has infinitely many normal modes.

The finite-dimensional modal ideas from M06-04 and M06-05 become a Fourier-series description of a continuous displacement field.

1 Small transverse displacement model

Consider a string whose equilibrium position lies along the

x
(6)

axis.

Let

y(x,t)
(7)

be its transverse displacement.

Assume:

  • the displacement is small,
  • the slope satisfies |∂y∕∂x|≪ 1,
  • the tension magnitude is nearly constant,
  • longitudinal motion is neglected,
  • the string is perfectly flexible.

These assumptions produce the linear string model.

2 Force balance on a small element

Take an element extending from

x
(8)

to

x + dx.
(9)

Its mass is

dm  = μ dx.
(10)

The tension at each end is tangent to the string.

Let the local tangent angle be

𝜃.
(11)

For small slope,

                ∂y
sin𝜃 ≈  tan 𝜃 ≈  --.
                ∂x
(12)

The net transverse force is therefore

dFy = T[ ∂y ||      ∂y || ]
 ---||     − ---||
 ∂x  x+dx   ∂x  x (13)
≈ T∂2y
--2-
∂xdx. (14)

Newton’s second law gives

    ∂2y-     ∂2y-
μ dx∂t2 =  T ∂x2dx.
(15)

Canceling

dx
(16)

gives

|--------------|
| ∂2y      ∂2y |
μ --2-=  T---2.|
--∂t------∂x----
(17)

PIC

Figure 1. A small string element experiences tension forces tangent to the string at its two ends. For small slopes, the difference of their transverse components is proportional to the local curvature.

3 Wave speed

Divide by

μ.
(18)

Then

∂2y    T ∂2y
----=  ------.
∂t2    μ ∂x2
(19)

Define

|--------|
| 2   T  |
|c =  μ-.|
---------
(20)

Thus

|----∘-----|
|       T  |
|c =    -. |
--------μ--|
(21)

The dimensions are

     ∘  ------
        --N---
[c] =   kg∕m  =  m∕s.
(22)

Larger tension increases the wave speed.

Larger linear mass density decreases it.

4 Traveling-wave solutions

A function of the form

|------------------|
-y(x,t) =-f(x-−-ct)-
(23)

satisfies the wave equation.

It represents a shape traveling to the right at speed

c.
(24)

Similarly,

|------------------|
-y(x,t) =-g(x-+-ct)|
(25)

travels to the left.

The general d’Alembert solution on an unbounded string is

|------------------------------|
|y(x,t) = f(x − ct) + g(x + ct).
--------------------------------
(26)

Standing Waves arise when oppositely traveling waves of the same frequency interfere.

5 A sinusoidal traveling wave

Consider

|----------------------------|
|y(x,t) = A cos(kx  − ωt + ϕ).|
------------------------------
(27)

Substitution into the wave equation gives

   2       2 2
− ω y = − c k y.
(28)

Therefore

|-------|
ω-=--ck-|
(29)

for positive

k.
(30)

The ideal string is nondispersive:

|------------|
|vp = ω-=  c,|
------k------|
(31)

and

|-------------|
|     dω-     |
vg =  dk = c. |
--------------
(32)

Unlike the discrete chain of M06-06, the ideal continuous string has a linear dispersion relation.

6 Standing wave from two traveling waves

Take equal-amplitude waves traveling in opposite directions:

y1 = A cos(kx − ωt),
(33)

and

y2 = − A cos(kx +  ωt).
(34)

Their sum is

y = A [cos(kx −  ωt) − cos(kx + ωt)].
(35)

Using a trigonometric identity,

|----------------------------|
|y(x,t) = 2A sin(kx) sin(ωt ). |
-----------------------------
(36)

The spatial shape

sin(kx)
(37)

does not translate.

Its amplitude oscillates in time.

This is a standing wave.

PIC

Figure 2. Two equal-frequency waves traveling in opposite directions can combine into a standing wave. The standing-wave nodes remain fixed in space while the displacement between them oscillates.

7 Fixed-end boundary conditions

Now consider a string extending from

x = 0
(38)

to

x =  L.
(39)

If both ends are fixed,

|----------|
y(0,t) = 0,|
------------
(40)

and

|------------|
|y(L, t) = 0. |
-------------
(41)

These boundary conditions select the allowed standing-wave patterns.

8 Separation of variables

Seek a product solution

|--------------------|
|y(x,t) = X (x)Q (t). |
---------------------
(42)

Substitution into

ytt = c2yxx
(43)

gives

X ¨Q = c2X ′′Q.
(44)

Divide by

c2XQ  :
(45)

 Q¨    X ′′
-2-- = ----.
c Q     X
(46)

The left side depends only on time.

The right side depends only on position.

Therefore both must equal a constant.

For oscillatory normal modes, write that constant as

− k2.
(47)

Then

|---------------|
| ′′   2        |
X---+-k-X--=-0,-
(48)

and

|--------------|
|¨     2       |
-Q-+-ω--Q-=--0,
(49)

where

|--------|
-ω-=-ck.-|
(50)

9 Solving the spatial equation

The spatial solution is

X (x) = A sinkx +  B coskx.
(51)

The boundary condition

X (0) = 0
(52)

requires

B  = 0.
(53)

Thus

X (x ) = A sin kx.
(54)

The second boundary condition,

X (L ) = 0,
(55)

requires

sin(kL ) = 0.
(56)

Therefore

|----------------------------|
|     nπ-                    |
|kn =  L ,    n =  1,2,3,....|
------------------------------
(57)

10 Normal frequencies

Since

ωn = ckn,
(58)

------------
|     n πc |
|ωn = ----.|
-------L----
(59)

The ordinary frequencies are

|---------------|
fn =  ωn-=  nc. |
------2π----2L---
(60)

Therefore

|----------|
-fn-=-nf1,-|
(61)

with

|--------------∘-----|
|     -c-   -1-   T- |
|f1 = 2L  = 2L    μ .|
---------------------|
(62)

The fixed-fixed ideal string has an exact harmonic spectrum.

11 Normal-mode shapes

Ignoring an arbitrary normalization constant,

|------------(----)--|
Xn (x ) = sin n-πx  .|
---------------L------
(63)

The first few modes are:

X1(x) = sin (πx-)
  L, (64)
X2(x) = sin (     )
  2πx-
   L, (65)
X3(x) = sin ( 3πx )
  ----
   L. (66)

PIC

Figure 3. The first three fixed-fixed string modes. Increasing mode number adds internal nodes and shortens the wavelength while preserving zero displacement at both endpoints.

12 Allowed wavelengths

For

k  =  nπ,
 n    L
(67)

and

      2π-
kn =  λn,
(68)

|----------|
|      2L  |
|λn =  ---.|
-------n---
(69)

Thus the string length contains

n
(70)

half-wavelengths:

|----------|
|      λn  |
|L = n ---.|
--------2--
(71)

13 Nodes and antinodes

A node is a point whose displacement remains zero throughout the normal-mode motion.

For mode

n,
(72)

nodes satisfy

   ( nπx )
sin  -L--  = 0.
(73)

Therefore

|----------------------------|
|     jL                     |
|xj = -n-,    j = 0,1, ...,n.|
------------------------------
(74)

The endpoints are always nodes.

Antinodes occur where the magnitude of the sine is one:

|---------------------|
|antinode   (2j +-1)L- |
xj      =     2n    , |
-----------------------
(75)

for the appropriate integer values of

j.
(76)

PIC

Figure 4. Nodes are fixed zeros of a standing-wave mode. Antinodes lie halfway between adjacent nodes and undergo the largest displacement magnitude.

14 Mode orthogonality

The fixed-end mode functions satisfy

|--------------------------------|
∫  L   ( nπx )    ( m πx)        |
|   sin  ----  sin   ----- dx =  0|
--0-------L----------L------------
(77)

for

n ⁄= m.
(78)

For

n = m,
(79)

|------------------------|
|∫ L     (     )         |
|    sin2  n-πx  dx =  L-.|
--0---------L---------2--
(80)

Combining both cases,

|----------------------------|
|∫ L                   L     |
|   Xn (x)Xm  (x)dx =  --δnm.|
--0--------------------2------
(81)

This is the continuum analogue of eigenvector orthogonality.

PIC

Figure 5. String normal modes act like orthogonal basis vectors in a function space. An arbitrary displacement shape can be projected onto them to obtain its modal amplitudes.

15 General modal expansion

Because the modes form a complete basis for suitable fixed-end shapes,

|------------------------------|
|         ∑∞          (     )  |
|y(x,t) =    Q  (t)sin  nπx-  .|
|              n         L     |
----------n=1-------------------
(82)

Substituting into the wave equation gives, mode by mode,

|----------------|
|¨Qn + ω2nQn =  0.|
-----------------
(83)

Thus

|--------------------------------|
|Qn (t) = An cosωnt + Bn sinωnt. |
---------------------------------
(84)

Therefore

y(x,t) = ∑ n=1∞[An cos ωnt + Bn sin ωnt] (85)
× sin ( nπx-)
   L. (86)

A continuous string is therefore an infinite collection of independent harmonic normal coordinates.

16 Initial displacement determines An

Suppose

|--------------|
y (x,0) = f(x).|
----------------
(87)

At

t = 0,
(88)

       ∑∞        (     )
f(x) =     A  sin  nπx-  .
             n      L
       n=1
(89)

Multiply by

sin (m πx ∕L )
(90)

and integrate.

Orthogonality gives

|------------------------------|
|      2 ∫ L        ( nπx )    |
An  = --    f (x)sin  ----  dx.|
------L---0------------L--------
(91)

These are Fourier sine coefficients.

17 Initial velocity determines Bn

Suppose

|---------------|
yt(x,0) = g(x). |
----------------
(92)

Differentiate the modal expansion:

          ∑∞           (n πx)
yt(x,0) =     ωnBn  sin  ----  .
          n=1             L
(93)

Project using orthogonality:

|---------------------------------|
|       2  ∫ L        ( nπx )     |
Bn  = ----     g(x)sin  ----  dx. |
------ωnL---0------------L---------
(94)

The complete initial state

f(x),g(x )
(95)

therefore determines every modal coordinate.

This is the continuous-field analogue of projecting finite-dimensional initial position and velocity onto Normal eigenvectors.

18 Example 1: fundamental frequency of a string

Let

L = 0.650 m, (96)
T = 120 N, (97)
μ = 5.00 × 10−3 kg∕m. (98)

The wave speed is

c = ∘ ------------
      120
  ---------−-3
  5.00 × 10 (99)
= √ ------
  24000 (100)
≈ 154.9 m∕s. (101)

The fundamental frequency is

f1 =  c
---
2L (102)
= 154.9-
1.300 (103)
≈ 119.2 Hz. (104)

The next two harmonics are

|--------------------------------|
f  ≈ 238.3 Hz,     f  ≈ 357.5 Hz.|
-2------------------3-------------
(105)

19 Example 2: string plucked into a triangular shape

Suppose the string is displaced to height

h
(106)

at

x = αL,
(107)

with

0 <  α < 1,
(108)

and is otherwise straight between the endpoints and the pluck point.

Thus

       (
       || hx-
       { αL ,        0 ≤ x ≤ αL,
f(x) =   h (L −  x)
       ||( ---------,  αL  ≤ x ≤ L.
         (1 − α)L
(109)

Assume the string is released from rest:

g(x) = 0.
(110)

Therefore

Bn =  0.
(111)

Evaluating the Fourier integral gives

|-----------------------------|
A   = -----2h-------sin (n πα ). |
| n   n2π2 α(1 − α)           |
-------------------------------
(112)

The motion is

y(x,t) = ∑ n=1∞A n cos(ωnt) (113)
× sin (     )
  nπx-
   L. (114)

The pluck position controls which harmonics are excited.

PIC

Figure 6. A triangular pluck is not a single normal mode. It is decomposed into an infinite Fourier sine series. The pluck position determines the relative amplitudes of the harmonics.

20 Midpoint pluck

For a midpoint pluck,

     1
α =  -.
     2
(115)

Then

|--------------(----)--|
|      -8h--     nπ-   |
-An-=--n2π2-sin---2---.|
(116)

For even

n,
(117)

sin(nπ ∕2) = 0.
(118)

Therefore

|--------|
|A2r = 0.|
----------
(119)

Only odd harmonics are excited by a perfectly symmetric midpoint pluck.

For the first few nonzero terms,

      8h-
A1 =  π2 ,
(120)

       -8h-
A3 = − 9π2 ,
(121)

and

       8h
A5 =  ---2-.
      25π
(122)

The

1∕n2
(123)

decay reflects the relatively smooth, piecewise-linear initial shape.

21 Nodes suppress selected modes

Suppose the string is plucked at a position

xp.
(124)

A mode whose node lies exactly at

x
 p
(125)

can have zero overlap with certain symmetric pluck profiles.

For the triangular result,

         (nπxp )
An ∝ sin  --L--  .
(126)

Thus if

|---(-----)------|
|sin  n-πxp  =  0,|
-------L----------
(127)

that harmonic is absent.

The geometry of the initial condition therefore selects the modal spectrum.

22 Kinetic energy of a string

The kinetic energy of an element

dx
(128)

is

             (   )
       1-      ∂y- 2
dK  =  2μ dx   ∂t   .
(129)

Thus

|--------∫-L-------|
|     1-      2    |
|K =  2μ  0  yt dx.|
--------------------
(130)

23 Elastic potential energy

For small slope, the extra length of a string element is approximately

           1 2
ds − dx ≈  -yxdx.
           2
(131)

Work against the tension gives

|--------∫---------|
|     1-   L  2    |
|U =  2T     yxdx. |
----------0---------
(132)

Therefore total energy is

|--------------------------|
|     1∫  L[          ]    |
|E =  --    μy2t + T y2x dx.|
------2--0------------------
(133)

24 Energy in modal coordinates

Insert

         ∑           ( nπx-)
y(x,t) =    Qn (t)sin   L    .
          n
(134)

Orthogonality eliminates cross terms.

The result is

|------------|
|    ∑∞      |
|E =     En, |
|     n=1    |
--------------
(135)

where

|------------------------|
|      μL [           ]  |
|En =  --- Q˙2n + ω2nQ2n  .|
--------4----------------
(136)

Thus each string normal mode behaves as an independent harmonic oscillator.

PIC

Figure 7. Orthogonality diagonalizes the string energy. Each modal coordinate Qn(t) carries its own conserved harmonic-oscillator energy in the ideal string model.

25 Effective modal mass

The modal energy can be written

En  = 1Mn  ˙Q2 +  1Mn ω2 Q2.
      2     n    2     n n
(137)

Comparing with the previous expression gives

|----------|
|      μL- |
|Mn  =  2  |
-----------
(138)

for the unnormalized choice

Xn =  sin(n πx∕L ).
(139)

The numerical modal mass depends on the normalization convention for

Xn.
(140)

The physical field

y(x,t)
(141)

does not.

26 Continuum orthogonality as mass orthogonality

For a nonuniform string with position-dependent density

μ(x ),
(142)

the natural modal inner product contains a Weight:

∫-----------------------|
|  μ(x)Xn (x)Xm  (x )dx. |
-------------------------
(143)

This is the continuous analogue of the finite-dimensional mass inner product

aT Mam.
  n
(144)

For a uniform string,

μ
(145)

is constant and can be factored out, leaving the ordinary sine orthogonality integral.

27 Fixed-free string

Now suppose

x = 0
(146)

is fixed but

x = L
(147)

is free in the idealized transverse sense.

The fixed end requires

|----------|
-X-(0) =-0.|
(148)

A free end has zero transverse force.

For small slope this gives

|----------|
X-′(L)-=-0.-
(149)

With

X  = A sinkx,
(150)

the free-end condition requires

cos(kL ) = 0.
(151)

Thus

|------------------------------------|
|      (2n −  1)π                     |
|kn =  ---------,     n = 1,2,3,.... |
----------2L-------------------------
(152)

Therefore

|----------------|
|fn = (2n-−-1-)c-.|
----------4L-----|
(153)

Only odd multiples of the fundamental occur.

28 Free-free string and the zero mode

For ideal free ends,

|--------------------------|
|X ′(0) = 0,     X ′(L ) = 0.|
---------------------------
(154)

There is a constant mode

X0 (x ) = constant,
(155)

with

|--------|
|ω0 = 0. |
---------
(156)

It corresponds to uniform transverse translation in the idealized mathematical model.

The remaining modes have

|--------------------------|
|     nπ-                  |
|kn =  L ,    n =  1,2,....|
----------------------------
(157)

Boundary conditions determine the spectrum just as constraints determine normal modes in finite-dimensional mechanics.

29 Discrete chain to continuous string

M06-06 derived the discrete chain equation

m ¨un =  κ(un+1 − 2un + un− 1).
(158)

For slow spatial variation,

un+1 − 2un +  un−1 ≈ a2uxx.
(159)

Thus

      κa2-
utt = m  uxx.
(160)

The string equation has

      T-
ytt = μ yxx.
(161)

Both have the same continuum form:

|----------------------------------------------------------------|
-inertia-density-×-acceleration-=-restoring-coefficient-×-curvature.-|
(162)

PIC

Figure 8. The oscillator-chain second difference becomes a spatial second derivative in the long-wavelength limit. Finite-dimensional normal coordinates become the standing-wave normal modes of a continuous string.

30 Why the continuous spectrum is still discrete on a finite string

The string has infinitely many degrees of freedom because

y(x,t)
(163)

is defined at continuously many positions.

Yet a finite string with boundary conditions has discrete allowed

kn.
(164)

The reason is not spatial discreteness.

It is the boundary-value eigenproblem.

Only certain spatial functions satisfy both the differential equation and the endpoint conditions.

Thus:

  • discrete lattices can have discrete or nearly continuous allowed k values depending on size and boundary conditions,
  • continuous media can have discrete normal frequencies when confined to finite domains.

31 Normal modes versus traveling waves

A normal mode of a finite fixed string is naturally represented as a standing wave.

A traveling wave is more natural on an unbounded string or before boundary reflections are combined.

The two descriptions are related.

A standing mode can be decomposed into equal-frequency left- and right-traveling waves.

Conversely, repeated reflection of traveling waves from boundaries builds the standing-wave normal-mode structure.

32 Common mistakes

  1. Using mass density per volume instead of linear mass density μ for the ideal string.
  2. Forgetting the small-slope approximation in the derivation.
  3. Treating the tension’s horizontal and transverse components as independently constant.
  4. Writing c = T∕μ instead of c = ∘ -----
  T ∕μ.
  5. Confusing traveling-wave speed with transverse material-point speed.
  6. Assuming a standing-wave pattern propagates along the string.
  7. Forgetting that fixed boundaries require zero displacement.
  8. Using kn = 2nπ∕L instead of nπ∕L for fixed-fixed modes.
  9. Confusing angular frequency ωn with ordinary frequency fn.
  10. Forgetting the factor of 2 in fn = nc∕(2L).
  11. Calling every Fourier component a distinct physical string rather than a mode of one string.
  12. Projecting initial displacement but forgetting initial velocity.
  13. Assuming all pluck positions excite all harmonics.
  14. Forgetting that modal mass depends on mode normalization.
  15. Applying fixed-fixed boundary conditions to fixed-free or free-free problems.

33 Practice exercises

  1. Derive the transverse force on a short string element and obtain the wave equation.
  2. Verify dimensionally that c = ∘  -----
   T∕μ has units of speed.
  3. Show directly that f(x − ct) satisfies the wave equation.
  4. Derive the standing-wave form from two counter-propagating sinusoidal waves.
  5. Use separation of variables to derive the fixed-fixed spatial and temporal equations.
  6. Derive kn = nπ∕L and fn = nc∕(2L).
  7. Find the nodes and antinodes of mode n.
  8. Prove sine-mode orthogonality over 0 ≤ x ≤ L.
  9. Derive the modal coefficient An from an arbitrary initial displacement.
  10. Derive the modal coefficient Bn from an arbitrary initial velocity.
  11. Derive the triangular-pluck coefficients.
  12. Show that a midpoint pluck excites only odd harmonics.
  13. Derive the total string energy and show that modal cross terms vanish.
  14. Derive the fixed-free spectrum.
  15. Explain how the discrete-chain continuum limit and the finite-string normal modes fit into one common normal-mode framework.

34 Summary

For a uniform stretched string,

|------------|
|μytt = T yxx,
--------------
(165)

or

|-------------------∘-----|
|                         |
ytt = c2yxx,    c =   T-. |
|                      μ  |
--------------------------
(166)

A fixed-fixed string has normal modes

|------------(n-πx)--|
Xn (x ) = sin ----  ,|
---------------L------
(167)

with

|----------|
|ωn = n-πc,|
-------L----
(168)

and

|---------|
|     nc- |
fn =  2L. |
-----------
(169)

An arbitrary fixed-end motion is

y(x,t) = ∑ n=1∞[A  cos ω t + B  sin ω  t]
  n     n     n     n (170)
× sin (     )
  nπx-
   L, (171)

where

|--------∫---------------------|
|     -2   L        ( nπx-)    |
An  = L     f (x)sin   L    dx,|
----------0---------------------
(172)

and

|----------∫-L--------(-----)-----|
B   = --2-     g(x)sin  nπx-  dx. |
| n   ωnL   0            L        |
-----------------------------------
(173)

The continuum field therefore decomposes into infinitely many independent harmonic normal coordinates.

This extends the normal-mode machinery from discrete mechanics directly into wave physics.

References

References

[1]   PhysicsLibrary, M06-07, Diatomic Oscillator Chains and Acoustic and Optical Branches.

[2]   PhysicsLibrary, M06-06, Discrete Oscillator Chains and Dispersion.

[3]   PhysicsLibrary, M06-05, Small Oscillations and the Normal-Mode Eigenvalue Problem.

[4]   A. P. French, Vibrations and Waves, W. W. Norton, 1971.

[5]   J. R. Taylor, Classical Mechanics, University Science Books, 2005.


"Continuous Strings, Standing Waves, and Normal Modes" is owned by bloftin.
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Also defines:  transverse string wave equation, standing wave, string normal mode, node, antinode, modal expansion, modal coordinate, fixed-end boundary condition
Keywords:  vibrating string, wave equation, standing waves, normal modes, Fourier series, boundary conditions, nodes, antinodes, string tension, linear mass density, modal coordinates, continuum limit

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