Question

A uniform rope of length 12m and mass 6kg hangs vertically from a rigid support. A block of mass 2kg is attached to the free end of the rope. A transverse pulse of wavelength $0.06\,m$ is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?
A) $0.06\,m$
B) $0.03\,m$
C) $0.12\,m$
D) $0.09\,m$

Answer 1

Hint : The wave travelling on the string will have different velocities depending on the tension in the rope. The lower point in the rope will have less tension since it supports less mass while the top of the rope will have more tension since it supports more mass.

Formula used: In this solution, will use the following solution
$\Rightarrow v = \sqrt {\dfrac{T}{\mu }}$ where $v$ is the velocity of the wave on the rope, $T$ is its tension, and $\mu$ is the linear mass density of the rope

Complete step by step answer
The wavelength of the wave travelling on the rope will change based on the tension in the string as the velocity of the wave changes with tension.
At the bottom of the rope, it will only have a tension corresponding to the 2 kg of mass of the block of mass. While at the top of the rope, it will be supporting a total 8 kg mass of the rope and the block together. From the relation, $v = \sqrt {\dfrac{T}{\mu }}$ , we can relate the velocity of the wave and the tension in the string at the lower and upper ends as
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}}$
Substituting the values of ${T_1} = 2g$ and ${T_2} = 8g$ , we can say
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{2g}}{{8g}}}$
$\Rightarrow {v_2} = 2{v_1}$
Since the frequency of the wave on the string will remain constant, from the relation $v = f\lambda$ , we can write
$\Rightarrow {\lambda _2} = 2{\lambda _1}$
Since the wavelength of the wave at the bottom of the rope is $0.06\,m$ , the wavelength at the top of the rope will be
${\lambda _2} = 0.12\,m$ which corresponds to option (C).

Note
Here we can directly use this relation only if the rope is uniform, that is its mass is uniformly distributed otherwise the wave of the string will be constantly changing and will depend on the distribution of mass in the string. The wavelength of the string will gradually increase as the wave moves from the bottom of the string to the top.