A uniform rope of length L and mass m_1 hangs vertically from a...

The velocity of a transverse wave on a string is given by $v = \sqrt{\dfrac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the mass per unit length.

Since the frequency $f$ of the wave remains constant as it travels up the rope, the wavelength $\lambda$ is proportional to the wave velocity $v$. Thus, $\lambda \propto \sqrt{T}$.

At the lower end of the rope, the tension is due to the attached block of mass $m_2$.

$T_1 = m_2 g$

At the upper end of the rope, the tension is due to the mass of the rope and the attached block.

$T_2 = (m_1 + m_2) g$

The ratio of the wavelengths is given by:

$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{T_2}{T_1}} = \sqrt{\dfrac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\dfrac{m_1 + m_2}{m_2}}$

Answer: $\sqrt{\dfrac{m_1 + m_2}{m_2}}$