Question

A uniform string of length 20 m 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
(take g = 10 m/s2)

Answer 1

Answer on Question #58904-Physics-Mechanics-Relativity

A uniform string of length $20\,\mathrm{m}$ 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 (take $g = 10\,\mathrm{m/s^2}$)

Solution

At any cords section at length $x$ above lowest point,

$$T = \frac{m g x}{l} = \mu g x$$

Hence speed of wave,

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{g x}$$

So,

$$\frac{d x}{d t} = \sqrt{g x} \rightarrow \int_{0}^{l} \frac{d x}{\sqrt{g x}} = \int_{0}^{t} d t$$ $$t = \frac{1}{\sqrt{g}} \left(\frac{x^{\frac{3}{2}}}{\frac{1}{2}}\right)_{0}^{l} = 2 \sqrt{\frac{l}{g}} = 2 \sqrt{\frac{20}{10}} = 2\sqrt{2}\,s.$$

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