Standing Waves, Resonance & Q (IPhO Waves)

IPhO waves lesson on standing waves and resonance: normal modes, boundary conditions, bandwidth, ring-down, and the quality factor Q.

  • International Physics Olympiad preparation
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Standing waves are what you get when a wave keeps reflecting and adding to itself in phase. Resonance is the driven version of the same idea: when the driving frequency matches a normal mode, energy piles up. The quality factor Q tells you how sharply tuned that pile-up is, and how slowly the oscillation dies away when you stop driving.

1. Definitions (Must Know)

  • Standing wave: superposition of two equal-frequency waves traveling in opposite directions.
  • Node / antinode: fixed zero-amplitude points and maximum-amplitude points of a standing wave.
  • Normal mode: a standing-wave pattern that satisfies the boundary conditions; labeled by an integer mode number.
  • Resonance: large steady-state response when a system is driven near a natural frequency.
  • Bandwidth (half-power): the frequency width Δ f between the two points where power is half the peak value.
  • Quality factor: Q = 2π (E/Δ E_cycle), and (for light damping) also Q = f₀/Δ f.
  • Ring-down: exponential decay after the drive is removed; amplitude envelope A(t) falls exponentially.

2. Key Ideas (What Earns Marks)

  • A standing-wave frequency is set by two things: wave speed v and a boundary-conditioned length scale L.
  • Start from boundary conditions, not memorized formulas, to avoid mixing up fixed, free, open, and closed ends.
  • For driven resonance, the high-scoring moves are: identify f₀, identify Δ f, then use Q = f₀/Δ f.
  • For ring-down data, extract the exponential decay constant from amplitude ratios, then connect to Q.
  • Always keep straight what is being measured: amplitude, intensity, power, or energy.

3. Detailed Explanations

3.1 Standing wave from two traveling waves

Add two waves of equal amplitude traveling in opposite directions:

y = A cos(kx-ω t) + A cos(kx + ω t) = 2A cos(kx) cos(ω t).

The spatial factor cos(kx) gives stationary nodes and antinodes, while the time factor cos(ω t) makes them oscillate.

3.2 Quantization of k from boundary conditions

For a string of length L:

  • Fixed-fixed: y(0,t) = 0 and y(L,t) = 0. A standing wave like y ∝ sin(kx) automatically satisfies y(0,t) = 0, and y(L,t) = 0 requires sin(kL) = 0:

    kₙ = nπ/L, fₙ = ωₙ/2π = nv/2L, n = 1,2,3,…
  • Fixed-free: y(0,t) = 0 and ∂ y/∂ x = 0 at x = L. Using y ∝ sin(kx), the slope condition is cos(kL) = 0, giving

    kₙ = ((2n-1)π)/2L, fₙ = ((2n-1)v)/4L, n = 1,2,3,…

For air columns, “closed end” behaves like a displacement node (particle velocity is zero), and “open end” behaves like a pressure node. The mode structure is the same logic: pick the sinusoid that satisfies the boundary conditions.

3.3 Driven resonance and bandwidth

A standard model is a driven, damped oscillator:

x double dot + 2βx dot + ω₀² x = F₀/m cos(ω t).

The steady-state amplitude is

A(ω) = (F₀/m)/(square root of ((ω₀²-ω²)² + (2βω)²)).

For light damping, the resonance is sharp and the half-power bandwidth in angular frequency is approximately

Δω ≈ 2β,

so

Q = ω₀/Δω = ω₀/2β.

Since ω = 2π f, the same ratio works with ordinary frequency: Q = f₀/Δ f.

3.4 Ring-down and the time meaning of Q

When the drive is removed, the underdamped solution has an envelope

A(t) = A₀ e^(-β t).

Energy scales like amplitude squared, so

E(t) = E₀ e^(-2β t).

Define the energy decay time constant τ_E by E(t) = E₀ e^(-t/τ_E), so τ_E = 1/(2β). Then

Q = ω₀/2β = ω₀τ_E.

A useful interpretation is the per-cycle energy loss fraction:

(Δ E_cycle)/E ≈ 2π/Q.

4. Common Mistakes

  • Using the fixed-fixed formula fₙ = nv/(2L) for open-closed pipes (only odd harmonics appear there).
  • Confusing amplitude half-maximum with power half-maximum when reading a resonance curve.
  • Mixing Δ f and Δω without the factor 2π.
  • Extracting β from energy decay but plugging it into an amplitude decay formula (or vice versa).
  • Forgetting that a mode number counts half-wavelengths along the length.

5. Exam Tips

  • Sketch the mode shape and mark nodes at the boundaries; it forces the correct sine or cosine choice.
  • If the question gives bandwidth, Q = f₀/Δ f is often the quickest mark.
  • If the question gives ring-down, take logs of amplitude ratios to isolate β cleanly.
  • If your computed resonance frequency is inconsistent with the boundary conditions (for example, a node where an antinode must be), restart from the boundary.

6. Worked Examples

1) Fixed-fixed string: first three resonant frequencies

A string has length L = 0.80 m and wave speed v = 120 m s⁻¹.

For fixed-fixed boundaries:

fₙ = nv/2L.

So

f₁ = 120/2(0.80) = 75 Hz, f₂ = 150 Hz, f₃ = 225 Hz.
2) Open-closed pipe: lowest three resonances

A pipe of length L = 0.50 m is closed at one end and open at the other. Take v = 340 m s⁻¹.

Open-closed modes have

fₙ = ((2n-1)v)/4L, n = 1,2,3,…

With 4L = 2.0 m,

f₁ = (1 · 340)/2.0 = 170 Hz, f₂ = (3 · 340)/2.0 = 510 Hz, f₃ = (5 · 340)/2.0 = 850 Hz.
3) Q from a resonance curve (half-power bandwidth)

A resonance peak is centered at f₀ = 1000 Hz. The half-power points are at 990 Hz and 1010 Hz.

Bandwidth:

Δ f = 1010-990 = 20 Hz.

Quality factor:

Q = f₀/(Δ f) = 1000/20 = 50.
4) Q from ring-down data

An oscillator has frequency f₀ = 200 Hz. After turning off the drive, the amplitude drops to 0.30 of its initial value after t = 2.0 s.

The envelope is A(t) = A₀ e^(-β t), so

A(t)/A₀ = e^(-β t) ⇒ β = 1/t ln(A₀/A(t)) = 1/2.0 ln(1/0.30) ≈ 0.60 s⁻¹.

Angular frequency ω₀ = 2π f₀ ≈ 2π(200) = 1.26 × 10³ rad s⁻¹.

For light damping, Q = ω₀/(2β):

Q ≈ (1.26 × 10³)/2(0.60) ≈ 1.0 × 10³.

7. Mind Stretchers

Mind-stretcher: relate cavity loss per round trip to Q

A 1D resonator of length L supports a mode near frequency f₀. Suppose the energy stored in the mode is multiplied by a factor g after each round trip due to losses, where g is slightly below 1.

Round-trip time:

Tᵣₜ = 2L/v.

If energy decays exponentially, E(t) = E₀ e^(-t/τ_E), then after one round trip:

g = E(Tᵣₜ)/E₀ = e^(-Tᵣₜ/τ_E) ⇒ τ_E = -Tᵣₜ/(ln g).

Since Q = ω₀τ_E with ω₀ = 2π f₀,

Q = -(2π f₀ Tᵣₜ)/(ln g).

This is a clean way to estimate Q from “fractional loss per bounce” models of strings, pipes, or optical cavities.

8. Practice

Practice

Quick drills:

  • Draw the first two mode shapes for fixed-fixed and fixed-free strings and label nodes.
  • Given f₀ and Δ f, compute Q and interpret it as “sharpness” of resonance.
  • Given two amplitude measurements in a ring-down, compute β using a log ratio.
Syllabus and review details

No official syllabus alignment is listed for this lesson.