Question

A uniform rope of length $12\text{ m}$ and mass $6\text{ kg}$ hangs vertically from a rigid support. A block of mass $2\text{ kg}$ is attached to the free end of the rope. A transverse pulse of wavelength $0.06\text{ m}$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is

• A. $0.8\text{ m}$
• B. $0.16\text{ m}$
• C. $0.12\text{ m}$
• D. $0.4\text{ m}$

Hint:

When a wave travels along a hanging system, its wavelength scales exactly with the square root of the supported mass ($\lambda \propto \sqrt{m_{\text{supported}}}$). Since the supported mass increases from $2\text{ kg}$ at the bottom to $8\text{ kg}$ at the top (a $4\times$ increase), the wavelength must increase by a factor of $\sqrt{4} = 2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A heavy, massive rope hangs vertically with a block attached to its bottom. A wave pulse travels up the rope. Because the rope has mass, the tension varies from the bottom to the top, which in turn alters the wave propagation speed and its wavelength.

Step 2: Key Formula or Approach:
The velocity $v$ of a transverse wave in a stretched string depends on the tension $T$ and mass per unit length $\mu$:
$$v = \sqrt{\frac{T}{\mu}}$$ Since the frequency $f$ remains constant as a wave travels through a medium, the velocity is directly proportional to wavelength ($v = f\lambda$). Therefore:
$$\lambda \propto v \propto \sqrt{T} \implies \frac{\lambda_{\text{top}}}{\lambda_{\text{bottom}}} = \sqrt{\frac{T_{\text{top}}}{T_{\text{bottom}}}}$$

Step 3: Detailed Explanation:
Let's calculate the tension at both reference points:
1. At the bottom end, the tension is created solely by the suspended block:
$$T_{\text{bottom}} = m_{\text{block}} \cdot g = 2g$$ 2. At the top end, the tension must support both the block and the entire weight of the rope:
$$T_{\text{top}} = (m_{\text{block}} + m_{\text{rope}}) \cdot g = (2 + 6)g = 8g$$ Now, set up the wavelength ratio using our proportionality relation:
$$\frac{\lambda_{\text{top}}}{0.06} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2$$ Isolating the top wavelength:
$$\lambda_{\text{top}} = 2 \times 0.06\text{ m} = 0.12\text{ m}$$

Step 4: Final Answer:
The wavelength at the top of the rope is $0.12\text{ m}$, which matches option (C).