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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
$\sqrt {\frac{{{m_1}}}{{{m_2}}}} $
$\sqrt {\frac{{{m_1} + {m_2}}}{{{m_2}}}} $
$\sqrt {\frac{{{m_2}}}{{{m_1}}}} $
$\sqrt {\frac{{{m_1} + {m_2}}}{{{m_1}}}} $
Solution
Speed of wave $\mathrm{V}=\sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$
$\lambda_{1}, \mathrm{V}_{1}$ are wavelength and speed of wave at bottom. $\lambda_{2}, \lambda_{2}$ are wavelength and speed of wave at top.
$\mathrm{V}=\mathrm{v} \lambda$
$\Rightarrow \lambda=\frac{V}{v}$
$\Rightarrow \lambda_{1}=\frac{V_{1}}{v} ; \lambda_{2}=\frac{V_{2}}{v}$
$\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\frac{V_{1}}{V_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}=\sqrt{\frac{m_{2} g}{\left(m_{1}+m_{2}\right) g}}$
$\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{m}_{2}}}$
