A string of length l is fixed at both ends and is vibrating in second harmonic. The amplitude at anti-node is 2 mm. The amplitude of a particle at distance `l//8` from the fixed end is

Options

A. `5 sqrt(2) mm`
B. `(5)/(sqrt(2))mm`
C. `5 mm`
D. `(10)/(sqrt(2))mm`

Explanation

To solve the problem, we need to determine the amplitude of a particle located at a distance \( \frac{L}{8} \) from one end of a string vibrating in its second harmonic. The amplitude at the antinode is given as 2 mm.

### Step-by-Step Solution:

1. **Understand the Harmonics**:
- The second harmonic of a string fixed at both ends has a node at the center and antinodes at the ends. The length of the string \( L \) is equal to one full wavelength \( \lambda \).
- Therefore, we have:
\[ \lambda = L \]

2. **Wave Equation**:
- The equation for the amplitude of a standing wave can be expressed as:
\[ y(x, t) = 2a \sin(kx) \cos(\omega t) \]
- Here, \( k \) is the wave number, and \( a \) is the amplitude at the antinode.

3. **Determine Wave Number**:
- The wave number \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{L} \]

4. **Amplitude Expression**:
- The amplitude \( A(x) \) at a distance \( x \) from one end is:
\[ A(x) = 2a \sin(kx) \]
- Substituting \( k \):
\[ A(x) = 2a \sin\left(\frac{2\pi}{L} x\right) \]

5. **Substituting Values**:
- We need to find the amplitude at \( x = \frac{L}{8} \):
\[ A\left(\frac{L}{8}\right) = 2a \sin\left(\frac{2\pi}{L} \cdot \frac{L}{8}\right) \]
- Simplifying this gives:
\[ A\left(\frac{L}{8}\right) = 2a \sin\left(\frac{\pi}{4}\right) \]

6. **Value of Sine**:
- We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
- Therefore:
\[ A\left(\frac{L}{8}\right) = 2a \cdot \frac{1}{\sqrt{2}} = \frac{2a}{\sqrt{2}} = a\sqrt{2} \]

7. **Finding \( a \)**:
- The amplitude at the antinode is given as \( 2 \, \text{mm} \). Thus, \( a = 2 \, \text{mm} \).
- Substituting this value:
\[ A\left(\frac{L}{8}\right) = 2 \sqrt{2} \, \text{mm} \]

8. **Final Calculation**:
- The amplitude at \( \frac{L}{8} \) is:
\[ A\left(\frac{L}{8}\right) = 2 \sqrt{2} \, \text{mm} \approx 2.83 \, \text{mm} \]

### Final Answer:
The amplitude of a particle at a distance \( \frac{L}{8} \) from the fixed end is approximately **2.83 mm**.