Question

A heavy but uniform rope of length L is suspended from a ceiling. (a) Write the velocity of a transverse wave travelling on the string as a function of the distance from the lower end. (b). If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the celiling ? (c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse ?

Answer 1

$m\to$ mass per unit length of string. Consider an element at distance 'x' from lower end.

Here wt acting down ward = (mx)g = Tension in the string of upper part velocity of transverse vibration =

$v=\sqrt{\left(\frac{T}{m}\right)}$

$=\sqrt{\left(\frac{mgx}{m}\right)}=\sqrt{\left(gx\right)}$

(b) For small displacement, $dt,\frac{dx}{\sqrt{\left(gx\right)}}$

Total time, $T={\int }_0^L\frac{dx}{\sqrt{\left(gx\right)}}=\sqrt{\left(\frac{4L}{g}\right)}$

(c) Suppose after time 't' from start the pulse meet the particle at distance 'y' from lower end y

$t={\int }_0^y\frac{dx}{\sqrt{\left(gx\right)}}$

$=\sqrt{\left(\frac{4y}{g}\right)}$

$\therefore$ Distance travelled by the particle in this time is (L - y).

$\therefore S=ut+\frac{1}{2}g{t}^2$

$\Rightarrow L-y=\left(\frac{1}{2}\right)g\times \left\{\right(\sqrt{\frac{4y}{g}}\left)\right\}$

$\Rightarrow L-y=2y$

$\Rightarrow 3y=L$

$\Rightarrow y=\frac{L}{3}$

So, the particle meet at distance $\frac{L}{3}$ from lower end.