A uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. A block of mass $m_2$ is…
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 $\lambda_2 / \lambda_1$ is [NEET 2016]
$\sqrt{\frac{m_1 + m_2}{m_2}}$
$\sqrt{\frac{m_2}{m_1}}$
$\sqrt{\frac{m_1 + m_2}{m_1}}$
$\sqrt{\frac{m_1}{m_2}}$
Solution
Wavelength of transverse pulse, $\lambda = \frac{v}{f}$ ...(i)
(where, $v = \text{velocity of the wave}$, $f = \text{frequency of the wave}$)
As, $v = \sqrt{\frac{T}{\propto}}$ ...(ii)
(where, $T = \text{tension in the rope}$, $\propto = \text{mass per unit length of the rope}$)
From Eqs. (i) and (ii), we get
$\lambda = \frac{1}{f} \sqrt{\frac{T}{\propto}} \Rightarrow \lambda \propto \sqrt{T}$
(A rope suspended vertically with mass $m_1$ attached above mass $m_2$.)
So, for two different cases, we get
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{m_1 + m_2}{m_2}}$