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

Answer 1

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

FBD of the string-block system



At point B, tension in the string $T_1=m_{2}g$

At point A, tension in the string $T_2=(m_1+m_2)g$

Velocity of a wave in a string is given by

$v=\sqrt{\frac{T}{\mu }}$

Mass density of the string $\mu$ is not changing.

Hence, ratio of velocities at B and A is

$\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{m_2}{m_1+m_2}}$

Since $v\times \lambda \left(\text{wavelength}\right)=f\left(\text{frequency}\right)$ which is constant for a wave,

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{v_2}{v_1}=\sqrt{\frac{m_1+m_2}{m_2}}$