Question

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 $\frac{\lambda _{2}}{\lambda _{1}}$ is:-

• A. $\sqrt{\frac{m_{1}}{m_{2}}}$
• B. $\sqrt{\frac{m_{1} + m_{2}}{ m_{2}}}$
• C. $\sqrt{\frac{\bar{m_{2}}}{m_{1}}}$
• D. $\sqrt{\frac{m_{1} + m_{2}}{m_{1}}}$

Answer 1

Answer: (B)

$V=\sqrt{\frac{T}{\mu }}\Rightarrow V_{1}=\sqrt{\frac{m_{2} g}{\mu }}$
$\Rightarrow V_{2}=\sqrt{\frac{\left(m_{1} + m_{2}\right) g}{\mu }}$
$V=f\lambda $
$\lambda \propto V$
$\frac{\lambda _{1}}{\lambda _{2}}=\frac{V_{1}}{ V_{2}}$
$\frac{\lambda _{2}}{\lambda _{1}}=\sqrt{\frac{m_{1} + m_{2}}{m_{2}}}$