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  • A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is:<div

A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is:

Option: 1

2 \pi \sqrt{2} s


Option: 2

2 s


Option: 3

2 \sqrt{2} s


Option: 4

\sqrt{2} s


Answers (1)

best_answer

The velocity of the wave at the height y of the string,

\frac{dy}{dt} = v = \sqrt{\frac{T}{\mu}}

        = \sqrt{\frac{\frac{\mu g y}{L}}{\frac{\mu}{L}}} = \sqrt {gy}

\therefore \frac{dy}{\sqrt y} = \sqrt{g}dt

\int_{0}^{L} \frac{dy}{\sqrt{y}} = \sqrt{g} \int_{0}^{t} dt

\implies 2 \sqrt{y} \int_{0}^{L}= \sqrt{g} t\int_{0}^{t}

\implies 2 \sqrt{L} = \sqrt{g} t

\implies t = 2 \sqrt{\frac{L}{g}} = 2 \sqrt{\frac{20}{10}} = 2 \sqrt{2} s

Posted by

shivangi.bhatnagar

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