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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=10 ms^-2)

Answer:

Explanation:

At distance x from the bottom

      $v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{(\frac{mgx}{L})}{(\frac{m}{L})}}=\sqrt{gx}$

    $\therefore    \frac{\text{d}x}{\text{d}t}=\sqrt{x}\sqrt{g}$

$\Rightarrow  \int_{0}^{L}x^{-\frac{1}{2}} dx=\sqrt{g}\int_{0}^{t} dt$

 $\Rightarrow [\frac{x^{-\frac{1}{2}}}{(\frac{1}{2})}\mid_0^L]=\sqrt{g}.t$

$\Rightarrow  t=\frac{2\sqrt{L}}{\sqrt{g}}$

$\Rightarrow  t=2\sqrt{\frac{20}{10}}=2\sqrt{2} s$