Wave Mechanics Examples: Boundary Conditions

This companion article provides exercises for WM12, wave mechanics: boundary Conditions. As in the other PhysicsLibrary exercise articles, all exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The central WM12 ideas are the distinction between initial conditions and boundary conditions, the two elementary endpoint conditions for an ideal string, and the way those conditions determine reflections and allowed normal modes.

For an ideal string, the two basic endpoint conditions are

|----------------| -fixed-end:-u-=-0--

(1)

and

|----------∂u------| |free end: ---= 0. | -----------∂x------|

(2)

At a fixed end, the reflected displacement is inverted. At a free end, the reflected displacement is not inverted [1, 3, 5].

For a string of length L with both ends fixed, the allowed standing-wave Wavenumbers, wavelengths, and frequencies are

|---------| |---------| |--------| |kn = n-π , |λn = 2L-, fn = nv-, -------L--| -------n--| ------2L--

(3)

with

n = 1,2, 3,...

(4)

For a string with one fixed end and one free end, the allowed values become

|---------------| |------------| |---------------| kn = (2n-−-1)π-, λn = --4L---, fn = (2n-−-1)v-, ---------2L------ ------2n-−-1-- ---------4L------

(5)

again with

n = 1,2, 3,...

(6)

Only odd multiples of the fundamental appear in the fixed-free case [1, 2, 4].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. Be careful to distinguish four ideas that students often mix together:

data at one time versus constraints at spatial edges,
zero displacement versus zero slope,
fixed-end reflection versus free-end reflection, and
fixed-fixed mode formulas versus fixed-free mode formulas.

Part I: Exercises

Exercise 1: Classify the given conditions

Consider a wave field u(x,t) on the interval 0 ≤ x ≤ L. Classify each statement below.

u(x, 0) = f(x)
(x, 0) = g(x)
u(0,t) = 0
(L,t) = 0
u(0,t) = a cos(ω_dt)

For each statement, say whether it is an initial condition or a boundary condition. For the boundary conditions, also say whether it is homogeneous or nonhomogeneous and whether it is of Dirichlet or Neumann type.

Exercise 2: Read the initial-boundary-value diagram

The following figure shows a rectangular x-t domain.

Figure. Initial data live along the line t = 0, while boundary data live along the edges x = 0 and x = L.

Use the figure to answer the following.

Along which line is the initial shape u(x, 0) = f(x) prescribed?
Along which line is the initial velocity (x, 0) = g(x) prescribed?
What boundary condition is shown at the left end?
What boundary condition is shown at the right end?
Explain in words why a complete wave problem generally needs both initial data and boundary data.

Exercise 3: Compare fixed-end and free-end reflection

The next figure compares reflections from a fixed end and a free end.

Figure. A crest approaches each boundary. The lower panels show the reflected pulse in each case.

Use the figure and the physical rules of WM12 to answer:

When a positive displacement crest reflects from a fixed end, what is the sign of the reflected pulse?
When the same crest reflects from a free end, what is the sign of the reflected pulse?
Which boundary condition explains why the fixed-end reflected pulse is inverted?
Which boundary condition explains why the free-end reflected pulse is not inverted?
If the incident pulse has peak displacement +5.0 mm, what is the reflected peak displacement in each case?

Exercise 4: Fixed-fixed normal modes

A string has length

L = 1.20m

(7)

and both ends are fixed. The wave speed is

v = 96 m/s.

(8)

Find the first three allowed wavelengths.
Find the first three allowed frequencies.
What is the spacing between adjacent allowed frequencies?
Is 200 Hz an allowed normal frequency for this ideal string?

Exercise 5: Fixed-free normal modes

A string has one fixed end and one ideal free end. Its length is

L = 0.80m

(9)

and its wave speed is

v = 80 m/s.

(10)

Find the first three allowed wavelengths.
Find the first three allowed frequencies.
Is 50 Hz an allowed normal frequency?
What is the next allowed frequency after the fundamental?
Explain why even multiples of the fundamental are absent.

Exercise 6: Read mode families from a diagram

The following figure compares the first few mode shapes for fixed-fixed and fixed-free systems.

Figure. The upper panel shows fixed-fixed modes. The lower panel shows fixed-free modes.

Answer the following.

Which system has nodes at both ends?
Which system has a node at one end and a slope-zero antinode at the other?
In the fixed-fixed fundamental, how many half-wavelengths fit into the length L?
In the fixed-free fundamental, how many quarter-wavelengths fit into the length L?
In the fixed-free case, why is the next allowed mode associated with 3λ∕4 rather than λ∕2?

Exercise 7: Derive the fixed-free wavenumber condition

Suppose a Standing Wave on 0 ≤ x ≤ L is written as

u (x,t) = B sin(kx )cos(ωt).

(11)

The left end is fixed and the right end is free.

Show that the fixed condition at x = 0 is automatically satisfied.
Differentiate u(x,t) with respect to x.
Apply the free-end condition at x = L.
Show that the allowed wavenumbers are
$(2n-−-1)π- kn = 2L .$ (12)
Convert this result into the wavelength formula.

Exercise 8: Driven and mixed boundary ideas

A finite string has boundary conditions

u(0,t) = 0

(13)

and

u (L,t) = 3.0 mm cos(40πt).

(14)

What type of boundary condition is present at x = 0?
What type of boundary condition is present at x = L?
Which of the two boundary conditions is nonhomogeneous?
What is the drive frequency in hertz?
Explain qualitatively why such a driven boundary could produce a large response when its frequency matches a natural frequency of the system.

Part II: Complete worked solutions

Solution 1: Classify the given conditions

$u(x,0) = f(x )$ (15)

is an initial condition because it specifies the state of the entire string at the single time t = 0.
$∂u ---(x,0) = g(x) ∂t$ (16)

is also an initial condition. It gives the initial velocity distribution at t = 0.
$u(0,t) = 0$ (17)

is a boundary condition. Because it prescribes the field value itself, it is a Dirichlet condition. Because the prescribed value is zero, it is homogeneous. Physically it represents a fixed end.
$∂u-(L, t) = 0 ∂x$ (18)

is a boundary condition. Because it prescribes a derivative, it is a Neumann condition. Because the right-hand side is zero, it is homogeneous. Physically it represents an ideal free end.
$u (0,t) = a cos(ωdt)$ (19)

is a boundary condition. It is again of Dirichlet type because it prescribes the field value. It is nonhomogeneous because the prescribed value is not zero. Physically it represents a driven boundary.

Solution 2: Read the initial-boundary-value diagram

The initial shape u(x, 0) = f(x) is prescribed along the line
$t = 0.$ (20)

This is the horizontal bottom edge of the domain.
The initial velocity
$∂u ---(x,0) = g(x) ∂t$ (21)

is also prescribed along the same line,

$t = 0.$ (22)
At the left boundary the figure shows
$u (0, t) = 0,$ (23)

which is a fixed-end condition.
At the right boundary the figure shows
$∂u ---(L, t) = 0, ∂x$ (24)

which is a free-end condition.
The wave equation determines the allowed local relation between space and time derivatives, but it does not by itself choose one unique motion. Initial data tell us how the motion starts, and boundary data tell us how the field must behave at the edges for all times. Both are needed to select the physical solution.

Solution 3: Compare fixed-end and free-end reflection

A positive crest reflecting from a fixed end returns with opposite sign. Therefore the reflected pulse is negative:
$|------------------------------| reflected sign at fixed end = − . --------------------------------$ (25)
At a free end the displacement is not inverted, so the reflected pulse remains positive:
$|------------------------------| |reflected sign at free end = +. | -------------------------------$ (26)
The fixed-end inversion is required by the boundary condition
$u = 0$ (27)

at the boundary. The incident and reflected displacements must cancel there.
The free-end noninversion is associated with the condition
$∂u --- = 0 ∂x$ (28)

at the boundary. The end is allowed to move, but its slope must vanish.
If the incident peak is
$+5.0 mm,$ (29)

then the reflected peak is

$|---------| −-5.0mm----$ (30)

for a fixed end and

$|---------| +5.0-mm----$ (31)

for a free end.

Solution 4: Fixed-fixed normal modes

For a fixed-fixed string,

2L nv λn = ---, fn = --. n 2L

(32)

With

L = 1.20 m, v = 96 m/s,

(33)

we compute the first three modes.

The allowed wavelengths are

λ₁	= = 2.40 m,	(34)
λ₂	= = 1.20 m,	(35)
λ₃	= = 0.80 m.	(36)

Therefore

|------------------------------------------| -λ1-=-2.40m,----λ2 =-1.20m,---λ3-=-0.80-m.-|

(37)

The allowed frequencies are

f₁	= = 40 Hz,	(38)
f₂	= = 80 Hz,	(39)
f₃	= = 120 Hz.	(40)

Thus

|f--=-40-Hz,---f-=--80Hz,---f--=-120-Hz.-| --1-------------2------------3-----------|

(41)

Adjacent allowed frequencies differ by the constant amount
$Δf = 40Hz.$ (42)
The allowed frequencies are integer multiples of 40 Hz. Since
$200 Hz = 5 × 40 Hz,$ (43)

it is allowed. It corresponds to

$n = 5.$ (44)

Therefore

$|--------------------------------------| -200Hz--is-an-allowed-normal--frequency.--$ (45)

Solution 5: Fixed-free normal modes

For a fixed-free string,

-4L---- (2n-−-1)v- λn = 2n − 1, fn = 4L .

(46)

With

L = 0.80 m, v = 80 m/s,

(47)

we obtain:

Wavelengths:

λ₁	= = 3.20 m,	(48)
λ₂	= = 1.07 m (approximately),	(49)
λ₃	= = 0.64 m.	(50)

Hence

|------------------------------------------| |λ1 = 3.20m, λ2 ≈ 1.07m, λ3 = 0.64 m. | -------------------------------------------

(51)

Frequencies:

f₁	= = 25 Hz,	(52)
f₂	= = 75 Hz,	(53)
f₃	= = 125 Hz.	(54)

Therefore

|----------------------------------------| |f1 = 25 Hz, f2 = 75Hz, f3 = 125 Hz. | -----------------------------------------

(55)

The number 50 Hz is not of the form
$(2n − 1) × 25 Hz.$ (56)

It is an even multiple of the fundamental, not an odd multiple. Therefore

$|--------------------| |50Hz is not allowed. | ----------------------$ (57)
The next allowed frequency after the fundamental 25 Hz is
$|------| |75Hz. | --------$ (58)
Even multiples are absent because the free-end slope condition and the fixed-end zero-displacement condition can be satisfied simultaneously only for odd quarter-wave patterns:
$λ 3λ 5λ L = -, ---, --,... 4 4 4$ (59)

These correspond to odd integers in the formula for k_n and therefore to odd multiples of the fundamental frequency.

Solution 6: Read mode families from a diagram

The fixed-fixed system has nodes at both ends.
The fixed-free system has a node at the fixed end and a slope-zero antinode at the free end.
In the fixed-fixed fundamental, one half-wavelength fits into the length L:
$|-------| L = λ. | -----2---$ (60)
In the fixed-free fundamental, one quarter-wavelength fits into the length L:
$|-------| | λ | L = -. | -----4---$ (61)
The pattern L = λ∕2 would place a node at both ends. That would violate the free-end slope condition at the right end. The next shape that satisfies a node at one end and zero slope at the other is
$|--------| | 3λ- | |L = 4 .| ----------$ (62)

This is why the next fixed-free mode corresponds to the next odd quarter-wave pattern rather than to λ∕2.

Solution 7: Derive the fixed-free wavenumber condition

We are given

u (x,t) = B sin(kx )cos(ωt).

(63)

At the fixed end,
$u(0,t) = B sin(0)cos(ωt) = 0.$ (64)

So the left boundary condition is automatically satisfied.
Differentiating with respect to x gives
$∂u ∂x- = Bk cos(kx)cos(ωt).$ (65)
At the free end x = L, the condition is
$∂u ---(L, t) = 0. ∂x$ (66)

Therefore

$Bk cos(kL )cos(ωt ) = 0.$ (67)

For a nontrivial standing wave, B≠0 and k≠0, so the boundary condition requires

$cos(kL ) = 0.$ (68)
The zeros of cosine occur at
$π 3π 5 π kL = --, ---,---,... 2 2 2$ (69)

This sequence can be written as

$kL = (2n-−-1-)π-, n = 1,2,3,... 2$ (70)

Hence

$|----------------| | (2n − 1)π | |kn = ----------.| ----------2L------$ (71)
Using
$2π- k = λ ,$ (72)

we obtain

$2π (2n − 1)π ---= ---------. λn 2L$ (73)

Solving for λ_n gives

$|-------------| λn = --4L--. | ------2n-−-1---$ (74)

Solution 8: Driven and mixed boundary ideas

The boundary conditions are

u(0,t) = 0

(75)

and

u (L,t) = 3.0 mm cos(40πt).

(76)

The condition at x = 0 is a homogeneous Dirichlet boundary condition. Physically it represents a fixed end.
The condition at x = L is a Dirichlet boundary condition as well, because the displacement value is prescribed directly. Physically it is a driven boundary.
The nonhomogeneous boundary is
$u (L,t) = 3.0 mm cos(40πt),$ (77)

because its prescribed value is not zero.
Comparing
$cos(40πt)$ (78)

with the standard form

$cos(ωt),$ (79)

we identify

$ω = 40π rad/s.$ (80)

Therefore the ordinary frequency is

$f = ω--= 40π-= 20 Hz. 2π 2π$ (81)

So

$f-=--20Hz.-| ------------$ (82)
A driven boundary continuously injects energy into the system. If the drive frequency matches or nearly matches one of the natural frequencies allowed by the other boundary conditions, successive oscillations reinforce the corresponding mode. This is Resonance, and it can produce a much larger steady-state response than a drive far from resonance [1, 2, 6].

Common mistakes

Mistake: saying that u(x, 0) = f(x) is a boundary condition. It is an initial condition.
Mistake: saying that a free end means zero displacement. In the ideal string model it means zero spatial slope.
Mistake: forgetting that a fixed-end reflection inverts the displacement pulse.
Mistake: applying the fixed-fixed frequency formula to a fixed-free system.
Mistake: assuming the fixed-free modes are all integer multiples of the fundamental. Only odd multiples appear.
Mistake: forgetting that a prescribed moving boundary such as u(L,t) = a cos(ωt) is nonhomogeneous.

What WM12E1 reinforces

This exercise set reinforces three major lessons of WM12.

First, initial conditions tell us how the motion starts, while boundary conditions tell us how the field must behave at the edges for all times.

Second, the two elementary endpoint conditions for a string,

|-------| |------| |∂u | -u-=-0-| and |∂x-= 0 , ---------

(83)

lead directly to different reflection rules.

Third, boundary conditions select the allowed global solutions. Fixed-fixed systems satisfy

|----------------------| k = nπ-, f = nv-, -n----L--------n----2L--

(84)

whereas fixed-free systems satisfy

|------------------------------------| | (2n-−-1)π- (2n-−--1)v-| |kn = 2L , fn = 4L .| --------------------------------------

(85)

This is why changing only one boundary changes the entire mode spectrum.

References

[1] A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W. Norton & Company, 1971.

[2] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

[3] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.5, “Interference of Waves,” including reflections from fixed and free boundaries.

[4] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.6, “Standing Waves and Resonance.”

[5] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, Volume I, Chapter 49, “Modes,” especially the sections on reflection and standing waves.

[6] Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves - The Physics of Waves, Fall 2016, MIT OpenCourseWare, Sections 5.1.2, 5.3.2, 5.4, and 5.5 on boundary conditions, fixed and free ends, and driven boundaries.

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