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. $\sqrt{\frac{m_2}{m_1}}$
• B. $\sqrt{\frac{m_1+m_2}{m_1}}$
• C. $\sqrt{\frac{m_1}{m_2}}$
• D. $\sqrt{\frac{m_1+m_2}{m_2}}$

Answer 1

The correct option is D $\sqrt{\frac{m_1+m_2}{m_2}}$


Let the tension in rope at lowest point be $T_1$ and that at highest point be $T_2$.
From F.B.D.

Hence,
$\Rightarrow T_1=m_{2}g$

$T_2=(m_1+m_2)g$
Velocity of pulse, $v=\sqrt{\frac{T}{\mu }}\Rightarrow v\propto \sqrt{T}$
For the same pulse ('$f$' is fixed).
$\Rightarrow v=f\lambda$
or $\lambda \propto v$
$\Rightarrow \lambda \propto \sqrt{T}$
Hence at lowest and highest point on rope, wavelength of pulse:
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_1}{T_2}}$ or, $\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_2}{T_1}}$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{(m_1+m_2)g}{m_{2}g}}=\sqrt{\frac{m_1+m_2}{m_2}}$