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)

    2\pi \sqrt{2}\: \: s

  • Option 2)

    2s

  • Option 3)

    2\sqrt{2}s

  • Option 4)

    \sqrt{2}s

 

Answers (1)

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)

2\pi \sqrt{2}\: \: s

This is an incorrect option.

Option 2)

2s

This is an incorrect option.

Option 3)

2\sqrt{2}s

This is the correct option.

Option 4)

\sqrt{2}s

This is an incorrect option.

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