Question
Two strings (A, B) having linear densities $\mu_{A} = 2 \times 10^{- 4}\text{ }kg/m$ and $\mu_{B} = 4 \times 10^{- 4}\text{ }kg/m$ and lengths $L_{A} = 2.5\text{ }m$ and $L_{B} = 1.5\text{ }m$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports C and D , respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time $t_{1}$ and $t_{2}$, respectively, to reach the joint. The ratio $t_{1}/t_{2}$ is :
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Solution
Given $L_{A} = 2.5\text{ }m$,
$$\begin{matrix} & L_{B} = 1.5\text{ }m \\ & \text{ }T = 500\text{ }N \\ & v_{A} = \sqrt{\frac{T}{\mu_{A}}} = \sqrt{\frac{500}{2 \times 10^{- 4}}} = 5\sqrt{10} \times 10^{2}\text{ }m/s \\ & v_{B} = \sqrt{\frac{T}{\mu_{B}}} = \sqrt{\frac{500}{4 \times 10^{- 4}}} = 5\sqrt{5} \times 10^{2}\text{ }m/s \\ & t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5\sqrt{10}} \times 10^{- 2}\text{ }s \\ & t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5\sqrt{5}} \times 10^{- 2}\text{ }s \\ & \ \therefore\ \frac{t_{1}}{t_{2}} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{5}{3} \times \frac{1}{\sqrt{2}} = \frac{1.66}{1.41} = 1.18 \end{matrix}$$
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