Physics - Wave on String Question with Solution | TestHub

The situation in the rope is given below,

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

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

${\mathit{v}}_1\mathit{=}\sqrt{\frac{{\mathit{m}}_2\mathit{g}L}{{\mathit{m}}_1}}$

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

${\mathrm{\lambda }}_1=\frac{{\mathrm{v}}_1}{\mathrm{f}}=\frac{1}{\mathrm{f}}\sqrt{\frac{{\mathrm{m}}_2\mathrm{g}\mathrm{L}}{{\mathrm{m}}_2}}$

Similarly, at top the velocity will be,

$v_1=\sqrt{\frac{(m_1+m_2)gL}{m_1}}$

And, the wavelength will be,

${\lambda }_2=\frac{1}{f}\sqrt{\frac{(m_1+m_2)gL}{m_1}}$

Hence, the ratio of both wavelengths will be,

$\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1}=\sqrt{\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}}$.