SATHEE: Wave Motion 1 Question 5

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, $g = 10 m s^{- 2}$ )

(2016 Main)

(a) $2 π \sqrt{2} s$

(b) $2 s$

(d) $\sqrt{2} s$

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Answer:

Correct Answer: 5. (c)

Solution:

At distance $x$ from the bottom

$\begin{aligned} v = & \sqrt{\frac{T}{μ}} & = \sqrt{\frac{(\frac{m g x}{L})}{(\frac{m}{L})}} = \sqrt{g x} \\ ∴ \frac{d x}{d t} & = \sqrt{x} \sqrt{g} \\ \Rightarrow & \int_{0}^{L} x^{- 1 / 2} d x & = \sqrt{g} \int_{0}^{t} d t \\ \Rightarrow [{\frac{x^{1 / 2}}{(1 / 2)} |}_{0}^{L}] & = \sqrt{g} \cdot t \\ \Rightarrow t & = \frac{2 \sqrt{L}}{\sqrt{g}} \\ \Rightarrow t & = 2 \sqrt{\frac{20}{10}} = 2 \sqrt{2} s \end{aligned}$

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Wave Motion 1 Question 5

5. 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 (Take, g=10ms−2 )

Answer:

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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, $g = 10 m s^{- 2}$ )