Question

A rope of mass m and length L hangs from a ceiling. a. Show that the wave speed on the rope a distance y above the lower end is v = 1gy. b. Show that the time for a pulse to travel the length of the string is ∆t = 22L/g.

Short answer

a: $v=\sqrt{gy}$

b: $∆t=2\sqrt{\frac{L}{g}}$

Step-by-step solution

Given Information

Mass of rope: m

Length of rope: L

y is an assumed length measured from bottom of the rope.

Part a:

We know that

$v=\sqrt{\frac{T_s}{\mu }}$

Where, v is speed of sound

$T_s$is tension in the string

and $\mu$is the linear density of the string.

$T_s=Mg$, where M is mass of string of y length.

g is acceleration due to gravity

Similarly $\mu =\frac{M}{y}$

Now substituting these values in equation (1) we get,

$$\begin{gathered} v=\sqrt{\frac{Mg}{{\displaystyle \frac{M}{y}}}} \\ \Rightarrow v=\sqrt{gy} \end{gathered}$$

Part b:

Now since we have the velocity as a function of length from bottom end of the rope, we can calculate time taken by wave to travel by integrating inverse of velocity over length L i.e.

$∆t={\int }_0^L\frac{dy}{v}$

$$\begin{gathered} ∆t={\int }_0^L\frac{dy}{v} \\ ={\int }_0^L\frac{dy}{\sqrt{gy}} \\ =\frac{1}{\sqrt{g}}{\int }_0^L\frac{dy}{\sqrt{y}} \\ =\frac{1}{\sqrt{g}}{\left[2\sqrt{y}\right]}_0^L \\ =2\sqrt{\frac{L}{G}} \end{gathered}$$