Question 7

**Setting up** Tension T is the same in all strings. Wave speeds: v₁ = √(T/μ), v₂ = √(T/4μ) = v₁/2, v₃ = √(T/16μ) = v₁/4. Since ω is the same everywhere, k₂ = ω/v₂ = 2k and k₃ = ω/v₃ = 4k. **At junction P (μ → 4μ, lighter to heavier)** - Reflection amplitude ratio: r = (v₂ − v₁)/(v₂ + v₁) = (v₁/2 − v₁)/(3v₁/2) = −1/3 → inverted (phase π), travelling in −x. Reflected wave: y = (1/3)y₀ cos(ωt + kx + π). **(A) correct**, α₁ = 1/3. - Transmission ratio: t = 2v₂/(v₁ + v₂) = 2/3. Transmitted wave in S₂ has wavenumber 2k: y = (2/3)y₀ cos(ωt − 2kx). Option (B) writes cos(ωt − kx) → **(B) incorrect**. **At junction Q (4μ → 16μ, again lighter to heavier)** - Reflected wave is inverted and travels in −x in S₂ with wavenumber 2k: y = α₃y₀ cos(ωt + 2kx + π). Option (C) has cos(ωt − kx + π) (wrong direction and wavenumber) → **(C) incorrect**. - Transmitted wave into S₃ travels in +x with wavenumber 4k, amplitude (2/3)·[2v₃/(v₂+v₃)] y₀ = (2/3)(2/3)y₀ = (4/9)y₀ > 0: y = (4/9)y₀ cos(ωt − 4kx). **(D) correct**, α₄ = 4/9. **Answer: (A) and (D)**

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