Question 164: Waves

Question

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 supports is :
(take ${\,\,g = 10m{s^{ - 2}}}$ )

Accepted Answer

We know that velocity in string is given by
$v = \sqrt {{T \over \mu }} \,\,\,\,\,\,\,\,\,\,\,\,....\left( i ight)$
where $\mu  = {m \over {m I}} = {{mass\,\,\,of\,\,\,string} \over {length\,\,\,of\,\,\,string}}$
The tension $T = {m \over \ell } 	imes x 	imes g\,\,\,\,\,\,\,\,\,\,...\left( {ii} ight)$
From $(a)$ and $(b)$ ${{dx} \over {dt}} = \sqrt {gx} $
${x^{ - 1/2}}\,dx = \sqrt g \,dt$
$	herefore$ $\int\limits_0^\ell  {{x^{ - 1/2}}} dx - \sqrt g \int\limits_0^\ell  {dt} $
$2\sqrt \ell   = \sqrt g  	imes t$
$	herefore$ $t = 2\sqrt {{\ell  \over g}}  = 2\sqrt {{{20} \over {10}}}  = 2\sqrt 2 $

Answer

$2\sqrt 2 s$

Answer

$2\pi \sqrt 2 s$

Answer

$2\pi \sqrt 2 s$

Answer

$2$ $s$

Waves

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 supports is : (take {\,\,g = 10m{s^{ - 2}}} )

Choose the correct answer: