Wave Equation & Boundary Conditions (IPhO Waves)

IPhO waves lesson on the 1D wave equation and boundary conditions: fixed/free ends, impedance, and reflection/transmission at interfaces.

International Physics Olympiad preparation

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Most “hard” wave problems are actually two easy steps:

write the right wave model in the bulk (usually the 1D wave equation), and
enforce the right boundary conditions (the part that decides phase flips, nodes, and energy flow).

1. Definitions (Must Know)

Wave field: a displacement or field component, written as y(x,t) in 1D.
1D wave equation: (∂² y)/(∂ t²) = v²(∂² y)/(∂ x²).
Wave speed (string): v = square root of (T/μ) where T is tension and μ is mass per unit length.
Harmonic wave: y(x,t) = A cos(kx-ω t + φ) with k = 2π/λ and ω = 2π f.
Dispersion relation (non-dispersive): ω = vk.
Mechanical impedance (string): Z = T/v = square root of Tμ (ratio of transverse force amplitude to transverse velocity amplitude for a traveling wave).
Fixed end boundary: y = 0 at the boundary.
Free end boundary: transverse force is zero, so ∂ y/∂ x = 0 at the boundary.
Interface boundary (two strings): y is continuous, and the transverse force T ∂ y/∂ x is continuous.

2. Key Ideas (What Earns Marks)

In the bulk, waves are usually superpositions of left- and right-traveling solutions.
Boundaries do not “add new physics”; they just enforce a condition on y and its derivatives.
A fixed end produces a phase inversion on reflection; a free end produces no inversion.
For interfaces, impedance controls reflection: a jump to higher impedance tends to invert the reflected wave.
Do not confuse amplitude coefficients (for y) with power coefficients (for energy flow).
Always check limiting cases: “interface becomes fixed end” and “interface becomes free end”.

3. Detailed Explanations

3.1 Deriving the 1D wave equation for a string

Take a short string element of length Δ x with tension T and small slope. The net vertical force is approximately

F_y ≈ T((∂ y)/(∂ x)|_(x + Δ x)-(∂ y)/(∂ x)|ₓ) ≈ T((∂² y)/(∂ x²))Δ x.

Newton’s second law gives

μΔ x (∂² y)/(∂ t²) = T((∂² y)/(∂ x²))Δ x ⇒ (∂² y)/(∂ t²) = (T/μ)(∂² y)/(∂ x²).

So v = square root of (T/μ).

3.2 General solution and the “two traveling waves” picture

For a non-dispersive 1D medium, the general solution can be written as

y(x,t) = f(x-vt) + g(x + vt),

which is a right-moving shape plus a left-moving shape. For single-frequency problems, it is usually faster to use complex amplitudes:

y(x,t) = Re {(Ae^ikx + Be^(-ikx))e^(-iω t)}, ω = vk.

3.3 Fixed end and free end reflections

Let the boundary be at x = 0, with incident amplitude A and reflected amplitude B (same k and ω).

Fixed end: y(0,t) = 0 implies A + B = 0, so B = -A and the reflection coefficient is

r = B/A = -1.

The reflected wave is inverted (a phase shift of π).
Free end: T ∂ y/∂ x = 0 implies ∂ y/∂ x = 0 at x = 0:

(∂ y)/(∂ x) ∝ ik(A-B) = 0 ⇒ B = A,

so

r = B/A = +1.

These two cases are also the infinite-impedance and zero-impedance limits of the general interface result below.

3.4 Reflection and transmission at an interface (impedance method)

Consider two semi-infinite strings joined at x = 0. A wave comes from medium 1 (impedance Z₁) toward medium 2 (impedance Z₂).

Boundary conditions:

continuity of displacement: Aᵢ + Aᵣ = Aₜ,
continuity of transverse force: Z₁(Aᵢ-Aᵣ) = Z₂ Aₜ (for harmonic waves at a fixed frequency).

Solving gives the amplitude coefficients (for displacement):

r = Aᵣ/Aᵢ = (Z₁-Z₂)/(Z₁ + Z₂), t = Aₜ/Aᵢ = 2Z₁/(Z₁ + Z₂).

The power coefficients use energy flux. For a traveling wave on a string, average power is proportional to Zω²A², so

R = Pᵣ/Pᵢ = r², T = Pₜ/Pᵢ = 4Z₁Z₂/((Z₁ + Z₂)²).

3.5 Quick translation to air columns

In pipes, the boundary condition is often written in terms of pressure p and particle velocity u:

closed end: u = 0 (displacement node),
open end: p ≈ 0 (pressure node). The same idea applies: decide what must vanish at the boundary, then build the standing wave pattern.

4. Common Mistakes

Using y = 0 for a free end (it is the slope or transverse force that is zero).
Forgetting that “inversion” is about the reflected wave’s phase, not the transmitted wave.
Mixing amplitude and power: R is not r; it is r².
Swapping Z₁ and Z₂ in the reflection coefficient.
Dropping the small-slope assumption when deriving the wave equation.

5. Exam Tips

Write the boundary condition in words first (fixed: displacement pinned; free: force zero), then translate to math.
Use limits as a self-check: if Z₂ is much larger than Z₁, the boundary should behave like a fixed end.
For interface problems, compute r first; t is often just 1 + r for displacement amplitude.
If the question asks about energy, finish with R and T, not just amplitudes.

6. Worked Examples

1) Fixed end reflection: find the reflected wave and the standing wave form

Incident wave:

yᵢ = A cos(kx-ω t).

At a fixed end at x = 0, the total displacement must satisfy y(0,t) = 0, so the reflected wave must have the same amplitude and opposite sign:

yᵣ = -A cos(kx + ω t).

The sum is a standing wave:

y = yᵢ + yᵣ = A[cos(kx-ω t)- cos(kx + ω t)] = 2A sin(kx) sin(ω t).

The boundary is a node because sin(0) = 0.

2) Free end reflection: show there is no inversion

For a free end at x = 0, ∂ y/∂ x = 0 at the end.

Take the same incident wave yᵢ = A cos(kx-ω t). A reflected wave with no inversion is

yᵣ = A cos(kx + ω t).

Then

y = yᵢ + yᵣ = 2A cos(kx) cos(ω t).

The boundary is an antinode because cos(0) = 1, and the slope vanishes because ∂(cos kx)/∂ x is zero at x = 0.

3) Interface: string joins a heavier string (same tension). Find r and transmitted power fraction

Let μ₂ = 4μ₁ and the tension is the same. Then

Z ∝ square root of μ ⇒ Z₂/Z₁ = square root of (μ₂/μ₁) = 2.

Reflection amplitude coefficient:

r = (Z₁-Z₂)/(Z₁ + Z₂) = (1-2)/(1 + 2) = -1/3.

So the reflected wave is inverted and has one-third the incident amplitude.

Power reflection coefficient:

R = r² = 1/9.

Power transmission coefficient:

T = 1-R = 8/9.

7. Mind Stretchers

Mind-stretcher: a string terminated by a mass has a frequency-dependent reflection coefficient

A string (impedance Z) is terminated at x = 0 by a small bead of mass m that can move transversely. A harmonic wave at angular frequency ω is incident from the string. Find the reflection coefficient r.

Let y(0,t) ∝ (Aᵢ + Aᵣ)e^(-iω t). The boundary condition is “net transverse force equals mass times acceleration”:

T(∂ y)/(∂ x)|₀ = m(∂² y)/(∂ t²)|₀.

For incident and reflected waves in the string,

(∂ y)/(∂ x)|₀ ∝ ik(Aᵢ-Aᵣ), (∂² y)/(∂ t²)|₀ ∝ -ω²(Aᵢ + Aᵣ).

Using Tk = ω Z and writing r = Aᵣ/Aᵢ, you get

r = (Z-i mω)/(Z + i mω).

Interpretation:

at low frequency, mω is small and r approaches + 1 (free-end-like),
at high frequency, the mass cannot accelerate easily and r approaches -1 (fixed-end-like).

8. Practice

Quick drills:

For a fixed end, write the reflected wave from yᵢ = A cos(kx-ω t) and simplify the sum into standing-wave form.
For an interface, compute r from impedances, then check the limits “heavier” and “lighter”.
Decide the correct boundary condition for a closed pipe end and an open pipe end.

Syllabus and review details

No official syllabus alignment is listed for this lesson.