A uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. A block of mass $m_2$ is at

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:

Accepted Answer

1. **Wave Speed on a String:** The speed $v$ of a transverse wave on a string under tension $T$ is given by $v = \sqrt{\frac{T}{\mu}}$, where $\mu$ is the linear mass density of the string .
2. **Relationship between Wavelength and Tension:** Since the frequency $f$ of the wave remains constant as it propagates, the wavelength $\lambda$ is directly proportional to the wave speed $v$ (because $v = f\lambda$). Thus, $\lambda \propto \sqrt{T}$.
3. **Tension at the Bottom ($T_1$):** At the lower end of the rope, the tension is entirely due to the weight of the attached block of mass $m_2$. Therefore, $T_1 = m_2g$.
4. **Tension at the Top ($T_2$):** At the upper end (at the rigid support), the tension must support the weight of both the rope and the attached block. Therefore, $T_2 = (m_1 + m_2)g$.
5. **Calculate the Ratio:** Using the proportionality $\lambda \propto \sqrt{T}$, the ratio of the wavelengths is:
   $$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2)g}{m_2g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$$

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