Two uniform strings, of linear mass densities μ and 4μ, are joined end to end and stretched under the same tension. A harmonic transverse wave of…

Wave speed is $v=\sqrt{T/\mu}$, so the speed in the heavier string is half that in the lighter one, $v_2=v_1/2$. Continuity of displacement and slope at the junction gives reflection and transmission amplitudes $r=\dfrac{v_2-v_1}{v_2+v_1}=-\dfrac13$ and $t=\dfrac{2v_2}{v_1+v_2}=\dfrac23$. So the reflected amplitude is $A_i/3$ (inverted), not $A_i/2$, and (A) is wrong. The transmitted amplitude is $\tfrac23A_i$, so (B) is correct. The power fraction reflected is $r^2=\tfrac19$, so (C) is correct. Frequency is the same on both sides, hence $\lambda=v/f$ halves in the heavier string, so (D) is correct. As a check, transmitted power fraction is $\tfrac{Z_2}{Z_1}t^2=2\times\tfrac49=\tfrac89$, and $\tfrac19+\tfrac89=1$. Answer: B, C and D.