Question

A uniform rope of mass m and length L hangs from a ceiling. (a) Show that the speed of a transverse wave on the rope is a function of y, the distance from the lower end, and is given by $v=\sqrt{gy}$. (b) Show that the time a transverse wave takes to travel the length of the rope is given by $t=2\sqrt{L/g}$.

Answer 1

we know, The wave speed at any point on the string is given by the equation

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

where: $\tau$ is not the tension along the whole string, it is the tension at the point where the speed need to be find but since we are usually dealing with uniform strings or ropes then $\tau$ is given for the whole string also,because in most cases the string is lying horizontally

we can write the tension as follows

$\tau =\frac{m}{L}gy=\mu gy$

By substitution to find the wave speed we have

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

b) For the second part, the time taken by the wave to travel from one end to the other end of the rope will be $dt=\frac{L}{v}$ and we noticed that we have used the differential representation for time because the speed $v$ is not the speed along the whole rope it is the speed at a particular point $y$ So now we need to integrate over $y$ to find $t$

$T={\int }_0^{L}dt={\int }_0^L\frac{L}{\sqrt{gy}}=2\sqrt{\frac{L}{g}}$