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JEE Advanced 2021 Paper 1 · Physics · Waves and Sound · medium

JEE Advanced 2021 Paper 1 Physics: Paragraph 2

Two strings are joined end to end and stretched under the same tension T=40 NT=40\ \mathrm{N}. The left string has linear mass density μ1=4.0 g m−1\mu_1=4.0\ \mathrm{g\,m^{-1}} and the right string has μ2=25 g m−1\mu_2=25\ \mathrm{g\,m^{-1}}. A harmonic transverse wave of frequency 100 Hz100\ \mathrm{Hz} and amplitude 3.0 mm3.0\ \mathrm{mm} travels along the left string towards the junction. At the junction the displacement and the slope of the string are continuous, so that part of the wave is reflected and part is transmitted. Take π=3.14159\pi=3.14159 and neglect any losses.

Question 1

Numerical answer

Find the amplitude, in millimetres, of the wave transmitted into the right string. Give the answer correct to two decimal places.
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Answer: 1.71

Continuity of displacement at the junction gives Ai+Ar=AtA_i+A_r=A_t. Continuity of slope gives k1(Ai−Ar)=k2Atk_1(A_i-A_r)=k_2A_t, with k=ω/vk=\omega/v. Hence Ai−Arv1=Atv2\frac{A_i-A_r}{v_1}=\frac{A_t}{v_2}. Solving, At=2v2v1+v2AiA_t=\frac{2v_2}{v_1+v_2}A_i and Ar=v2−v1v1+v2AiA_r=\frac{v_2-v_1}{v_1+v_2}A_i. The wave speeds are v1=T/μ1=40/0.004=100 m s−1v_1=\sqrt{T/\mu_1}=\sqrt{40/0.004}=100\ \mathrm{m\,s^{-1}} and v2=40/0.025=40 m s−1v_2=\sqrt{40/0.025}=40\ \mathrm{m\,s^{-1}}. Therefore At=2×40140×3.0=1.714A_t=\frac{2\times40}{140}\times3.0=1.714 mm, i.e. 1.711.71 mm, and Ar=−60140×3.0=−1.29A_r=-\frac{60}{140}\times3.0=-1.29 mm (a phase reversal on reflection from the denser string).

Question 2

Numerical answer

Find the time-averaged power, in milliwatts, carried away from the junction by the transmitted wave in the right string. Give the answer correct to two decimal places.
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Answer: 580.09

The average power of a travelling wave on a string is P=12μvω2A2=12Tμ ω2A2P=\tfrac12\mu v\omega^2A^2=\tfrac12\sqrt{T\mu}\,\omega^2A^2. Here ω=2π×100=628.32 s−1\omega=2\pi\times100=628.32\ \mathrm{s^{-1}}. In the right string, Tμ2=40×0.025=1.0 kg s−1\sqrt{T\mu_2}=\sqrt{40\times0.025}=1.0\ \mathrm{kg\,s^{-1}} and At=80140×3.0×10−3=1.7143×10−3A_t=\frac{80}{140}\times3.0\times10^{-3}=1.7143\times10^{-3} m. So Pt=12×1.0×(628.32)2×(1.7143×10−3)2=0.5801P_t=\tfrac12\times1.0\times(628.32)^2\times(1.7143\times10^{-3})^2=0.5801 W =580.09=580.09 mW. Check by energy conservation: the incident power is 1240×0.004 ω2(3×10−3)2=710.61\tfrac12\sqrt{40\times0.004}\,\omega^2(3\times10^{-3})^2=710.61 mW, the reflected power is 130.52130.52 mW, and 710.61−130.52=580.09710.61-130.52=580.09 mW, which agrees. The transmitted power fraction is 4Z1Z2/(Z1+Z2)2=0.8164Z_1Z_2/(Z_1+Z_2)^2=0.816.
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