Wave Motion 1 Question 5
5. 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, $g=10 ms^{-2}$ )
(2016 Main)
(a) $2 \pi \sqrt{2} s$
(b) $2 s$
(c) $2 \sqrt{2} s$
(d) $\sqrt{2} s$
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Answer:
Correct Answer: 5. (c)
Solution:
- At distance $x$ from the bottom
$$ \begin{array}{rlrl} v= & \sqrt{\frac{T}{\mu}} & =\sqrt{\frac{\left(\frac{m g x}{L}\right)}{\left(\frac{m}{L}\right)}}=\sqrt{g x} \\ & \therefore \quad \frac{d x}{d t} & =\sqrt{x} \sqrt{g} \\ \Rightarrow \quad & \quad \int _0^{L} x^{-1 / 2} d x & =\sqrt{g} \int _0^{t} d t \\ \Rightarrow \quad\left[\left.\frac{x^{1 / 2}}{(1 / 2)}\right| _0 ^{L}\right] & =\sqrt{g} \cdot t \\ \Rightarrow \quad t & & =\frac{2 \sqrt{L}}{\sqrt{g}} \\ \Rightarrow \quad t & =2 \sqrt{\frac{20}{10}}=2 \sqrt{2} s \end{array} $$
