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

Step 1: (a) The velocity of a transverse wave traveling on the string as a function of the distance from the lower end.

Mass per unit of length of string be $m$

Tension in the string be $T$

and acceleration due to gravity be $g$

Considering an element at distance $x$ from the lower end.

Here weight acting downward $=$Tension in the string of upper part $=\left(mx\right)g$

Thus, the velocity of transverse vibration $=v=\sqrt{(T/m)}=\sqrt{(mxg/m)}=\sqrt{\mathbf{g}\mathbf{x}}$

Step 2: (b) Time is taken by the pulse to reach the ceiling from the bottom:

Let the total distance from bottom to ceiling be $L$

For small displacement $dx,$ $dt=\frac{dx}{\sqrt{gx}}$…………..$\left(1\right)$

Integrating on both the sides we get,

Total time, ${\int }_0^{T}dt=\frac{1}{\sqrt{g}}{\int }_0^L\frac{dx}{\sqrt{x}}$

$\mathit{T}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{4}\mathbf{L}}{\mathbf{g}}}$ …………(Here $T$ is the total time required.)

Step 3: (c) Distance at which particle meets the pulse:

Suppose, it will meet the pulse after $y$ distance.

To get the in-between time, we integrate the equation $\left(1\right)$

${\int }_0^{t}dt=\frac{1}{\sqrt{g}}{\int }_0^y\frac{dx}{\sqrt{x}}$

$t=\sqrt{\frac{4y}{g}}$

Therefore, the distance traveled by the particle in this time is$(L-y)$

We know the relation,$S=ut+\frac{1}{2}g{t}^2$

$\Rightarrow L-y=\frac{1}{2}g{\left(\sqrt{\frac{4y}{g}}\right)}^2$ at $u=0$

$\begin{array}{rcl}& \Rightarrow & L-y=2y\\ & \Rightarrow & \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{=}\frac{\mathbf{L}}{\mathbf{3}}\end{array}$

Thus, the particle meets the pulse at distance $\frac{L}{3}$ from the lower end.