Two strings are joined at x = 0 and maintained under the same tension. Their linear densities are μ₁ = μ and μ₂ = 4μ. A wave is incident from string 1 and is represented by yᵢ = A cos(kx-ωt). Which pair correctly gives the reflected wave in string 1 and the transmitted wave in string 2?
Options

Question

Two strings are joined at x = 0 and maintained under the same tension. Their linear densities are μ₁ = μ and μ₂ = 4μ. A wave is incident from string 1 and is represented by yᵢ = A cos(kx-ωt). Which pair correctly gives the reflected wave in string 1 and the transmitted wave in string 2?

Options

G1 Physics · Accepted Answer

Because μ₂ = 4μ₁, the impedances satisfy Z₂ = 2Z₁. Hence, Aᵣ/Aᵢ = (Z₁−Z₂)/(Z₁+Z₂) = (1−2)/(1+2) = −1/3. Therefore, Aᵣ = −A/3. Also, Aₜ/Aᵢ = 2Z₁/(Z₁+Z₂) = 2/3, so Aₜ = 2A/3. The frequency does not change at the junction. Since v = √(T/μ), the transmitted speed is v₂ = v₁/2. Therefore, k₂ = ω/v₂ = 2k. Thus, yᵣ = −(A/3) cos(kx+ωt), yₜ = (2A/3) cos(2kx−ωt). Correct option: A.