Waves Part 2

Concept:The speed of a transverse wave on a string depends only on tension and linear density; time taken is length divided by speed.Explanation:Wave speed on a string: $v = \sqrt{\frac{T}{\mu}}$ where $T$ is tension, $\mu$ is linear density.Time to travel a length $L$ is $t = \frac{L}{v} = L \sqrt{\frac{\mu}{T}}$.Tension $T = 500\,\text{N}$ is same in both strings because they are joined and under same support force.For string A: $L_A = 2.5\,\text{m}$, $\mu_A = 2 \times 10^{-4}\,\text{kg/m}$$t_1 = 2.5 \sqrt{\frac{2 \times 10^{-4}}{500}}$For string B: $L_B = 1.5\,\text{m}$, $\mu_B = 4 \times 10^{-4}\,\text{kg/m}$$t_2 = 1.5 \sqrt{\frac{4 \times 10^{-4}}{500}}$Ratio: $\frac{t_1}{t_2} = \frac{2.5}{1.5} \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}} = \frac{5}{3} \sqrt{\frac{1}{2}} = \frac{5}{3\sqrt{2}}$Numerically: $\frac{5}{3\sqrt{2}} \approx \frac{5}{3 \times 1.414} \approx \frac{5}{4.242} \approx 1.18$Answer:Option B (1.18)