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.
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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:
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:
The steady-state amplitude is
For light damping, the resonance is sharp and the half-power bandwidth in angular frequency is approximately
so
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
Energy scales like amplitude squared, so
Define the energy decay time constant τ_E by E(t) = E₀ e^(-t/τ_E), so τ_E = 1/(2β). Then
A useful interpretation is the per-cycle energy loss fraction:
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:
So
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
With 4L = 2.0 m,
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:
Quality factor:
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
Angular frequency ω₀ = 2π f₀ ≈ 2π(200) = 1.26 × 10³ rad s⁻¹.
For light damping, Q = ω₀/(2β):
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:
If energy decays exponentially, E(t) = E₀ e^(-t/τ_E), then after one round trip:
Since Q = ω₀τ_E with ω₀ = 2π f₀,
This is a clean way to estimate Q from “fractional loss per bounce” models of strings, pipes, or optical cavities.
8. 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.