Question

A uniform rope of length $L$ and mass ${m_1}$ hangs vertically from a rigid support. A block of mass ${m_2}$ is attached to the free end of the rope. A transverse pulse of wavelength ${\lambda _1}$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is ${\lambda _2}$. The ratio is ${\lambda _2}/{\lambda _1}$:
A. $\sqrt {\dfrac{{{m_1}}}{{{m_2}}}}$
B. $\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_2}}}}$
C. $\sqrt {\dfrac{{{m_2}}}{{{m_1}}}}$
D. $\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_1}}}}$

Answer 1

Use the formula for the speed of a wave on the rope in terms of tension in the rope and also the formula for the speed of a wave in terms of wavelength of the wave. Combine these two equations to derive the relation between the wavelength of the wave and tension in the rope. Calculate the values of tensions in the rope using the concept of Newton’s second law of motion.

Formulae used:
The speed $v$ of a wave on a rope is
$v = \sqrt {\dfrac{T}{\mu }}$ …… (1)
Here, $T$ is the tension in the rope and $\mu$ is the linear density of the rope.
The speed $v$ of a wave is given by
$v = n\lambda$ …… (2)
Here, $n$ is the frequency of the wave and $\lambda$ is the wavelength of the wave.

Complete step by step answer:
We have given that a rope of length $L$ and mass ${m_1}$ is attached vertically to a rigid support and a block of mass ${m_2}$ is attached at the end of the rope. The wavelength of the wave at the end of the rope is ${\lambda _1}$ and the wavelength when it reaches the top of rope is ${\lambda _2}$.
There is a tension ${T_1}$ in the rope when the wave at the end of the rope is balanced by the weight ${m_2}g$ of the rope.
${T_1} = {m_2}g$
There is a tension ${T_2}$ in the rope when the wave reaches at the top of the rope is balanced by the weight $\left( {{m_1} + {m_2}} \right)g$ of the rope and the block.
${T_2} = \left( {{m_1} + {m_2}} \right)g$
From equation (1), we can conclude that the speed of a wave on the rope is directly proportional to the tension in the rope.
$v \propto \sqrt T$
From equation (2), we can conclude that the speed of the wave is directly proportional to the wavelength of the wave.
$v \propto \lambda$
From these two equations, we can write
$\lambda \propto \sqrt T$
Hence, the wavelengths ${\lambda _1}$ and ${\lambda _2}$ of the wave at the lower end and top of the rope is
$\dfrac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}}$
Substitute $\left( {{m_1} + {m_2}} \right)g$ for ${T_2}$ and ${m_2}g$ for ${T_1}$ in the above equation.
$\dfrac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\dfrac{{\left( {{m_1} + {m_2}} \right)g}}{{{m_2}g}}}$
$\therefore \dfrac{{{\lambda _2}}}{{{\lambda _1}}} = \sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_2}}}}$
Therefore, the required ratio of wavelengths is $\sqrt {\dfrac{{{m_1} + {m_2}}}{{{m_2}}}}$.

Hence, the correct option is B.

Note: The students may take the values of the two tensions in the rope wrongly. The tension ${T_1}$ is balanced by only the weight of the block as the wave is initially at the lower end of the rope where the block is attached and the tension ${T_2}$ is balanced by the weight of the block as well as rope because the wave is at the top of the rope.