Section 5 — Superposition of Collinear SHMs & Beats
GE-4 / PHYS7021 unit 5 · 2 lecture-hours · 3 solved questions
5.1 Theory you need
y1 = A1sin ωt, y2 = A2sin(ωt+φ) ⇒ y = R sin(ωt+θ), R = √(A12+A22+2A1A2cos φ), tan θ = A2sin φ/(A1+A2cos φ)Beats: two waves of slightly different frequencies f1, f2 give a waxing–waning sound; beat frequency = |f1−f2| per second.
5.2 Solved past questions — exam-style answers
1. Find the resultant amplitude and phase of y1 = 3sin(50πt) cm and y2 = √3 cos(50πt) cm.
Given: A1 = 3, and y2 = √3 cos(50πt) = √3 sin(50πt + π/2) ⇒ A2 = √3, φ = π/2.
- R = √(A12 + A22 + 2A1A2cos φ) = √(9 + 3 + 0) = √12 = 2√3 cm.
- tan θ = A2sin φ/(A1+A2cos φ) = √3/3 = 1/√3 ⇒ θ = π/6 = 30°.
- Resultant: y = 2√3 sin(50πt + π/6) cm.
∴ R = 2√3 cm, phase = π/6 (30°)
2. What are beats? How are they applied in the determination of poisonous gases in mines?
- Beats: when two sound waves of nearly equal frequencies superpose, the loudness at a point rises and falls periodically; one rise + one fall = one beat; beat frequency = f1 − f2 per second.
- Mine application: two identical organ pipes are sounded — one in pure air, one in the mine air.
- Poisonous gases (firedamp, CO) change the density of the air mixture ⇒ speed of sound and hence the pipe's frequency change.
- Counting the beats per second gives the frequency shift, which indicates the presence/amount of the poisonous gas.
∴ Beats = |f₁−f₂| per second; gas detection by beat count
3. Sonometer numerical: tensions, lengths, diameters, densities in ratios 8:1, 36:35, 4:1, 1:2; higher pitch = 360 Hz. Beats produced?
Formula: f = (1/Ld)√(T/ρ) × constant, since mass per unit length μ ∝ ρd2.
- f1/f2 = (L2/L1)(d2/d1)·√[(T1/T2)(ρ2/ρ1)] = (35/36)(1/4)·√(8×2).
- = (35/144) × 4 = 35/36.
- Higher frequency f2 = 360 Hz ⇒ f1 = 350 Hz.
- Beats = 360 − 350 = 10 per second.
∴ 10 beats per second