Question

A uniform string of length $20 \,m$ is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is : (take $g\, = \,10\,ms^{-2}$)

• A. $2 \pi \sqrt{2} s $
• B. $2s$
• C. $2 \sqrt{2} s $
• D. $ \sqrt{2} s $

Answer 1

The Correct Option is C

The problem requires determining the time taken by a wave pulse to travel up a uniform string suspended from a rigid support. Let's break down the solution step-by-step:

1. **Understanding the Problem:**
   • The string is uniform and suspended, meaning the tension in the string varies with height due to gravity.
   • A wave pulse traveling along the string will be affected by the varying tension.
2. **Formula for Wave Speed on a String:**
   • The wave speed \( v \) on a string is given by \( v = \sqrt{\frac{T}{\mu}} \), where:
     • \( T \) is the tension in the string at a point,
     • \( \mu \) is the linear mass density of the string.
3. **Calculate Tension:**
   • Consider a small element of the string at a distance \( x \) from the bottom.
     • The tension at a distance \( x \) is due to the weight of the string below it.
     • Therefore, \( T = \mu g x \), where \( g = 10 \, \text{m/s}^2 \).
4. **Expression for Wave Speed:**
   • Substitute \( T = \mu g x \) into the wave speed formula:
     v = \sqrt{\frac{\mu g x}{\mu}} = \sqrt{g x}
5. **Calculate Time Taken (Integration):**
   • Time for the pulse to travel the entire length \( L = 20 \, \text{m} \) is given by integrating the inverse of speed:
     t = \int_{0}^{L} \frac{1}{\sqrt{g x}} \, dx
   • Evaluate the integral:
     t = \int_{0}^{L} \frac{1}{\sqrt{10 x}} \, dx = \frac{1}{\sqrt{10}} \int_{0}^{20} \frac{1}{\sqrt{x}} \, dx
     = \frac{1}{\sqrt{10}} \left[ 2\sqrt{x} \right]_{0}^{20}
     = \frac{2}{\sqrt{10}} [\sqrt{20} - 0]
     = \frac{2 \times \sqrt{20}}{\sqrt{10}}
     = \frac{2 \times \sqrt{4 \times 5}}{\sqrt{10}}
     = \frac{2 \times 2 \times \sqrt{5}}{\sqrt{10}}
     = \frac{4 \times \sqrt{5}}{\sqrt{10}}
     = 2 \sqrt{2} \, \text{s}
6. **Conclusion:**
   • The time taken for the wave pulse to reach the top of the string is 2 \sqrt{2} \, \text{s}.
   • The correct option is therefore 2 \sqrt{2} \, \text{s}.