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$\mathit{v}\mathbf{=}\sqrt{\mathbf{g}\mathbf{y}}$ .(b)Show that the time a transverse wave takes to travel the length of the rope is given by$\mathit{t}\mathbf{=}\mathbf{2}\sqrt{\mathbf{L}\mathbf{/}\mathbf{g}}$.

Short answer

1. The speed of a transverse wave on the rope is a function of y, the distance from the lower end is given by $v=\sqrt{\left(gy\right)}$
2. The time a transverse wave takes to travel the length of the rope is given by$t=2\sqrt{\left(\frac{L}{g}\right)}$

Step-by-step solution

The given data

• The mass of the rope is m.
• Length of the rope is L.

Understanding the concept of the wave equation

The wave speed in a stretched string will be reciprocal of the square root of linear density of the string, provided tension in the string should be unity.

Formula:

Wave speed of a given string, $V=\sqrt{\frac{T}{\mu }}$ (i)

The linear density of a string,$\mu =\frac{m}{L}$ (ii)

a) Proving that the speed of the wave is a function of distance, y

Using equation (ii), the mass of the string is given as:

$m=\mu L$

Therefore, the mass of the string below point y with length y can be given as,

$m_y=\mu y$

Tension in the string below point y is

localid="1660980314188" $T=\mu yg\left(\because \mathrm{Gravitational}\mathrm{force}\mathrm{balances}\mathrm{the}\mathrm{tension}\mathrm{force}\mathrm{in}\mathrm{the}\mathrm{string}\right)$

Using value of we get

$T=\mu yg.............................................\left(1\right)$

Using equation (1) in equation (i), we get the speed of the wave as:

$$\begin{gathered} v=\sqrt{\frac{\mu yg}{\mu }} \\ T=\mu yg.............................................\left(2\right) \end{gathered}$$

Hence, it is proved that the wave speed is a function of distance, y.

b) Proving that the time taken to travel the length of the rope is t=2L/g

Integrating equation 2, we get

$$\begin{gathered} v=\sqrt{gy} \\ \frac{dy}{dt}=g^{\frac{1}{2}}\times y^{\frac{1}{2}} \\ \frac{1}{g^{\frac{1}{2}}}\times y^{-\frac{1}{2}}dy=dt \\ \frac{1}{\sqrt{g}}\left(y_{}^{-\frac{1}{2}}\right)dy=dt \end{gathered}$$

Integrating the left hand side from role="math" localid="1660979848533" $y=0toy$and the right hand side from$t=0toT$

$$\begin{gathered} \frac{1}{\sqrt{g}}{\int }_0^L\left(y^{-\frac{1}{2}}\right)dy={\int }_0^{T}dt \\ \frac{2}{\sqrt{g}}{\left[y^{-\frac{1}{2}}\right]}_0^L={\left[t\right]}_0^T \\ \frac{2}{\sqrt{g}}\sqrt{L}=T \\ T=2\sqrt{\frac{L}{g}} \end{gathered}$$

Hence, it is proved that the value of time taken is $T=2\sqrt{\frac{L}{g}}$.