A uniform string of length `20m` 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 = 10ms^(-2)`)

Options

A. 2s
B. `2sqrt(2) s`
C. `sqrt(2)s`
D. `2pi sqrt(2)s`

Answer

To solve the problem of determining the time taken for a wave pulse to travel up a uniform string of length 20 m, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform string of length \( L = 20 \, \text{m} \) suspended from a rigid support.
- A wave pulse is introduced at the lower end of the string.

2. **Identifying the Variables**:
- Let \( \mu \) be the mass per unit length of the string.
- The tension \( T \) in the string at a distance \( x \) from the bottom can be expressed as \( T = \mu g x \), where \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity.

3. **Wave Speed on the String**:
- The speed of the wave pulse \( v \) on the string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu g x}{\mu}} = \sqrt{g x} \]

4. **Relating Speed to Distance and Time**:
- The wave speed can also be expressed as:
\[ v = \frac{dx}{dt} \]
- Therefore, we can equate the two expressions for speed:
\[ \frac{dx}{dt} = \sqrt{g x} \]

5. **Separating Variables**:
- Rearranging gives:
\[ dt = \frac{dx}{\sqrt{g x}} \]

6. **Integrating**:
- To find the total time \( T \) taken for the wave to travel from \( x = 0 \) to \( x = L \):
\[ T = \int_0^L \frac{dx}{\sqrt{g x}} \]
- This integral can be solved as follows:
\[ T = \int_0^{20} \frac{dx}{\sqrt{10 x}} = \frac{1}{\sqrt{10}} \int_0^{20} x^{-1/2} \, dx \]
- The integral of \( x^{-1/2} \) is \( 2\sqrt{x} \):
\[ T = \frac{1}{\sqrt{10}} \left[ 2\sqrt{x} \right]_0^{20} = \frac{1}{\sqrt{10}} \left( 2\sqrt{20} - 0 \right) = \frac{2\sqrt{20}}{\sqrt{10}} = 2\sqrt{2} \]

7. **Final Result**:
- Thus, the time taken for the wave pulse to reach the support is:
\[ T = 2\sqrt{2} \, \text{seconds} \]