Lesson 13: Waves — Continued

Week 13 — Standing waves, sound intensity, and review

🎯 Learning Objectives

Analyze standing waves and harmonics on strings and in pipes
Calculate sound intensity levels and power
Solve multi-concept wave problems

📋 Standing Waves in Pipes

Sound waves in tubes also form standing waves:

Open-open pipe: λₙ = 2L/n, fₙ = nv/2L

Open-closed pipe: λₙ = 4L/n, fₙ = nv/4L (n = 1, 3, 5, ... only odd harmonics)

📋 Sound Intensity

Sound intensity decreases with distance (spherical spreading):

I = P / (4πr²) (point source radiating equally in all directions)

Decibel scale: β = 10 log₁₀(I/I₀), where I₀ = 10⁻¹² W/m²

📝 Worked Example: Intensity at Different Distances

A speaker produces sound with intensity 0.5 W/m² at 2 m. What is the intensity at 5 m? What is the decibel level at 5 m?

Step 1 — Intensity ratio:

I₁/I₂ = r₂²/r₁²

0.5/I₂ = 5²/2² = 25/4 = 6.25

I₂ = 0.5/6.25 = 0.08 W/m²

Step 2 — Decibel level at 5 m:

β = 10 log₁₀(0.08/10⁻¹²) = 10 log₁₀(8×10¹⁰) = 10 × 10.90 = 109 dB

✅ I₂ = 0.08 W/m², β = 109 dB (very loud — near the threshold of pain!)

📋 Wave Energy

Energy of a wave ∝ A²f² (amplitude squared × frequency squared)

Doubling the amplitude quadruples the energy. This is why a slightly louder sound (more amplitude) carries significantly more energy.

✅ Check Your Understanding

Q1: Why does an open-closed pipe only have odd harmonics?

Click to reveal answer

The open end must be an antinode and the closed end must be a node. The distance between a node and the nearest antinode is λ/4. So the fundamental fits λ/4 into the pipe: L = λ/4, giving λ = 4L. The next mode fits 3λ/4: L = 3λ/4, giving λ = 4L/3. Then L = 5λ/4, etc. Only odd multiples work: λ = 4L/n where n = 1, 3, 5, ...

Q2 (mini-problem): If the distance from a point source doubles, by what factor does the intensity decrease?