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 ceiling? (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

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as 

\[v = \sqrt{\left( \frac{T}{m} \right)} = \sqrt{\left( \frac{mgx}{m} \right)}\] 

\[ \Rightarrow v = \sqrt{\left( gx \right)}\]
(b) Let the time taken be dt for the small displacement dx.
Thus, we have:
\[dt = \frac{dx}{v} = \frac{dx}{\sqrt{\left( gx \right)}}\]
\[\therefore \text{ Total  time, }   T =  \int\limits_0^L \frac{dx}{\sqrt{\left( gx \right)}} = \sqrt{\left( \frac{4L}{g} \right)}\]
(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.
Now,

\[t =  \int\limits_0^y \frac{dx}{\sqrt{\left( gx \right)}}\] 

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

∴ Distance travelled by the particle in this time, S = \[L - y\]
Using the equation of motion, we get:

\[S = ut + \frac{1}{2}  g t^2 \] 

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

\[ \Rightarrow L - y = 2y\] 

\[ \Rightarrow 3y = L\] 

\[ \Rightarrow y = \frac{L}{3}\]
Thus, the particle will meet the pulse at a distance
\[\frac{L}{3}\] from the lower end.