A wire of length 2L is made by joining two wi | vedclass

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(A) Let the mass per unit length of wires $A$ and $B$ be $\mu_1$ and $\mu_2$ respectively.Since the materials are the same,the density $\rho$ is the s

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$1 : 2$

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(A) Let the mass per unit length of wires $A$ and $B$ be $\mu_1$ and $\mu_2$ respectively.Since the materials are the same,the density $\rho$ is the same.$\mu_1 = \rho \pi r^2$ and $\mu_2 = \rho \pi (2r)^2 = 4 \rho \pi r^2 = 4 \mu_1$.The tension $T$ is the same in both wires.The wave speed in the wires is $v = \sqrt{T/\mu}$.Thus,$v_1 = \sqrt{T/\mu_1} = v$ and $v_2 = \sqrt{T/\mu_2} = \sqrt{T/(4\mu_1)} = v/2$.For a wire of length $L$ fixed at both ends,the frequency of the $n$-th harmonic is $f = n \frac{v}{2L}$.Since the joint is a node,both wires vibrate independently with the same frequency $f$.$f = p \frac{v_1}{2L} = q \frac{v_2}{2L}$,where $p$ and $q$ are the number of loops (which equals the number of antinodes).$p \frac{v}{2L} = q \frac{v/2}{2L} \implies p = q/2 \implies p/q = 1/2$.Therefore,the ratio $p : q$ is $1 : 2$.

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