# 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) Option 1) Option 2) Option 3) Option 4)

As we learnt in

Simple Harmonic Motion -

It is special case of oscillatory motion of a vibrating particle in which the acceleration at any position is directly propotional to displacement from mean position & it is directed towards mean position.

- wherein

$\vec{a}\; \alpha -\vec{x}$

At any cords section at length x above lowest point.

$T=\frac{mgx}{l}=\mu gx$            $\because\ \; \frac{m}{l}=\mu$

We know that $v=\sqrt{\frac{T}{\mu}}=\sqrt{gx}$

$\frac{dx}{dt}=\sqrt{gx}\ \; \Rightarrow\ \; \int_{0}^{l}\frac{dx}{\sqrt{x}}=\int_{0}^{l}dt$

$t=\frac{1}{\sqrt{g}}\left(\frac{x^{1/2}}{1/2} \right )_{0}^{l}\ \; \Rightarrow\ \; 2\sqrt{\frac{l}{g}}\ \; \Rightarrow\ \; 2\sqrt{\frac{20}{10}}=2\sqrt{2}\ sec$

Correct option is 3.

Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is the correct option.

Option 4)

This is an incorrect option.

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