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 λ 1 is produced at th

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 λ 1 ​ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ 2 ​ . The ratio λ 1 ​ λ 2 ​ ​ is

TestHub · Accepted Answer

The velocity of a transverse wave on a string is given by $v = \frac{T}{μ}$ , where $T$ is the tension and $μ$ is the mass per unit length.

Since the frequency $f$ of the wave remains constant as it travels up the rope, the wavelength $λ$ is proportional to the wave velocity $v$ . Thus, $λ \propto T$ .

At the lower end of the rope, the tension is due to the attached block of mass $m_{2}$ .

$T_{1} = m_{2} g$

At the upper end of the rope, the tension is due to the mass of the rope and the attached block.

$T_{2} = (m_{1} + m_{2}) g$

The ratio of the wavelengths is given by:

$\frac{λ _{2}}{λ _{1}} = \frac{T _{2}}{T _{1}} = \frac{( m _{1} + m _{2} ) g}{m _{2} g} = \frac{m _{1} + m _{2}}{m _{2}}$

Answer: $\frac{m _{1} + m _{2}}{m _{2}}$

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