# A uniform thin rope of length $12\; m$ and mass $6\; kg$ has vertically support and a block of mass $2\; kg$ is attached to its free end. A transverse short wave train of wavelength 6cm is produced at the lower end of the rope. What is the wavelength of the wave train (in cm) when it reaches top of the rope     Option: 1 3 Option: 2 6 Option: 3 12   Option: 4 9

$\mu =\frac{6}{12}=0.5$

using $V=\frac{T}{\mu}$

At a distance x from the bottom

$V=\sqrt{\frac{T}{\mu}}= \sqrt{\frac{(2g+\mu xg)}{\mu}}=\sqrt{\frac{20+5x}{0.5}}$

at x=0, $V_1=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{20}{0.5}}=\sqrt{40}$

and $\lambda _1=6 \ cm$

$f=\frac{V_1}{\lambda _1}=\frac{\sqrt{40}}{6}$

Now at x=12 cm

$V_0=\sqrt{\frac{T}{\mu}}= \sqrt{\frac{(2g+\mu xg)}{\mu}}=\sqrt{\frac{20+5x}{0.5}}=\sqrt{\frac{20+(5*12)}{0.5}}=\sqrt{\frac{80}{0.5}}\\ \Rightarrow V_0=2\sqrt{40}$

So, $\lambda _0=\frac{V_0}{f}=\frac{2\sqrt{40}}{\frac{\sqrt{40}}{6}}=12 cm$

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