Question

A uniform string of length $20\text{m}$ 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 $g=10{\text{ms}}^{-2}$)

• A. $2\pi \sqrt{2}\text{s}$
• B. $2\text{s}$
• C. $2\sqrt{2}\text{s}$
• D. $\sqrt{2}\text{s}$

Answer 1

The correct option is C $2\sqrt{2}\text{s}$

A uniform string of length $20\text{m}$ is suspended from a rigid support. Let $m$ be the total mass of string.

Then, tension in the string at distance $x$ from bottom,

$T=\frac{mgx}{l}$

So, velocity at point $P$, $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{\frac{mgx}{l}}{m/l}}$

i.e. $v=\sqrt{gx}$

Comparing with kinematic equation, $v^2=u^2+2as$, pulse acts like a particle starting from rest at the bottom and travelling with uniform acceleration $g/2$ upwards.

Hence, time taken for pulse to reach the top,
$t=\sqrt{\frac{2h}{g/2}}=\sqrt{\frac{2\times 20}{5}}=2\sqrt{2}\text{s}$

Alternative method:

$v=\frac{dx}{dt}=\sqrt{gx}$

$\Rightarrow {\int }_0^{20}\frac{dx}{\sqrt{x}}={\int }_0^t\sqrt{g}dt$

$\Rightarrow [2\sqrt{x}{]}_0^{20}=\sqrt{10}t\Rightarrow 2\sqrt{20}=\sqrt{10}t$

or $t=2\sqrt{2}s$