A uniform rope of length L and mass m _ 1 , hangs vertically from a...

The situation in the rope is given below,

The velocity of the wave at the bottom of the rope is given by,

$v = \sqrt{\dfrac{T}{\mu}}$; where $T$ is tension and $\mu$ is linear mass density of the string.

${v}_{1} = \sqrt{\dfrac{{m}_{2}gL}{{m}_{1}}}$

Hence, the wavelength for frequency $f$ will be,

${\lambda}_{1} = \dfrac{{v}_{1}}{f} = \dfrac{1}{f}\sqrt{\dfrac{{m}_{2}gL}{{m}_{2}}}$

Similarly, at top the velocity will be,

${v}_{1} = \sqrt{\dfrac{( {m}_{1} + {m}_{2} )gL}{{m}_{1}}}$

And, the wavelength will be,

${\lambda}_{2} = \dfrac{1}{f}\sqrt{\dfrac{( {m}_{1} + {m}_{2} )gL}{{m}_{1}}}$

Hence, the ratio of both wavelengths will be,

$\dfrac{{\lambda}_{2}}{{\lambda}_{1}} = \sqrt{\dfrac{{m}_{1} + {m}_{2}}{{m}_{2}}}$.