A string of length 0.4, m and mass 10^{-2}, k | vedclass

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(B) The linear mass density of the string is $\mu = \frac{M}{L} = \frac{10^{-2}}{0.4} = 2.5 	imes 10^{-2}\, kg/m$.The velocity of the wave pulse in t

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$0.10$

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(B) The linear mass density of the string is $\mu = \frac{M}{L} = \frac{10^{-2}}{0.4} = 2.5 	imes 10^{-2}\, kg/m$.The velocity of the wave pulse in the string is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1.6}{2.5 	imes 10^{-2}}} = \sqrt{\frac{160}{2.5}} = \sqrt{64} = 8\, m/s$.For constructive interference,the pulse must return to the starting point in the same phase. When a pulse reflects from a fixed end,it undergoes a phase change of $\pi$. After two reflections (traveling a total distance of $2L$),the pulse undergoes a total phase change of $2\pi$,returning to its original phase.Therefore,the time taken for the pulse to travel $2L$ is $\Delta t_{\min} = \frac{2L}{v} = \frac{2 	imes 0.4}{8} = \frac{0.8}{8} = 0.1\, s$.

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