Question

A transverse pulse generated at the bottom of the uniform vertical rope of length $L$ travels in upward direction. The time taken by the pulse to reach the top of the rope will be :

• A. $\sqrt{\frac{L}{2g}}$
• B. $\sqrt{\frac{2L}{g}}$
• C. $\sqrt{\frac{L}{g}}$
• D. $\sqrt{\frac{4L}{g}}$

Answer 1

The correct option is D $\sqrt{\frac{4L}{g}}$

Velocity of the transverse wave is given by,
$v=\sqrt{\frac{T}{\mu }}$
The mass per unit length of rope is,
$\mu =\frac{m}{L}$
The tension in rope at a distance ${}^{\prime }{x}^{\prime }$ from bottom will be,
$T_x=W_x$
$\Rightarrow T_x=\left(\frac{m}{L}\right)xg=\mu xg$
Therefore the velocity of wave pulse at a distance $x$ from the bottom will be,
$v_x=\sqrt{\frac{T_x}{\mu }}$
$\Rightarrow v_x=\sqrt{\frac{\mu xg}{\mu }}$
$\Rightarrow v_x=\sqrt{xg}$
The velocity of wave pulse is varying with the distance from bottom, hence time taken by pulse to travel through a very small distance $dx$ will be:
$dt=\frac{dx}{v_x}$
(Considering that velocity of pulse will not change during such a small distance)
$\Rightarrow dt=\frac{dx}{\sqrt{gx}}....\left(i\right)$
Now integrating Eq.$\left(i\right)$ with proper limits for the wave pulse reaching at the top of rope:
$(x=0)\to (x=L)$
and $t=0\text{to}t=t$
$\Rightarrow {\int }_0^{t}dt=\frac{1}{\sqrt{g}}{\int }_0^L\frac{1}{\sqrt{x}}dx$
$\Rightarrow t=\frac{1}{\sqrt{g}}{\left[2\sqrt{x}\right]}_0^L$
$\Rightarrow t=2\sqrt{\frac{L}{g}}$
$\therefore t=\sqrt{\frac{4L}{g}}$
$\Rightarrow \left(d\right)$ is correct option.