Question

A heavy uniform rope hangs vertically from the ceiling, with its lower end free. A disturbance on the rope travelling upwards from the lower end has a velocity \(v\) at a distance \(x\) from the lower end such that :

• A. \( v \propto x \)
• B. \( v \propto \sqrt{x} \)
• C. \( v \propto \frac{1}{x} \)
• D. \( v \propto \frac{1}{\sqrt{x}} \)

Hint:

In hanging rope, tension increases with depth, so wave speed increases upward.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The velocity of a transverse wave in a rope depends on the tension at that point and the mass per unit length. In a hanging rope, tension is due to the weight of the part below the point.
: Key Formula or Approach:
1. \( v = \sqrt{\frac{T}{\mu}} \) where \( \mu \) is mass per unit length.
2. Tension \( T \) at distance \( x \) from bottom = weight of length \( x \).
Step 2: Detailed Explanation:
Let \( \mu \) be the mass per unit length of the rope.
Mass of length \( x \) from the lower end \( = \mu x \).
Weight of this section (which is the Tension at distance \( x \)):
\[ T = (\mu x) \cdot g \]
Velocity of wave:
\[ v = \sqrt{\frac{T}{\mu}} \]
\[ v = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{gx} \]
Therefore:
\[ v \propto \sqrt{x} \]
Step 3: Final Answer:
The velocity is proportional to the square root of distance from the bottom (\( v \propto \sqrt{x} \)).