A streteched string of length l fixed at both ends can sustain stationary waves of wavelength `lambda` given by

Options

A. `lambda = n^(2)//2l`
B. `lambda = l^(2)//2 n`
C. `lambda = 2l//n`
D. `lambda = 2ln`

Explanation

To solve the problem of a stretched string of length \( l \) fixed at both ends that can sustain stationary waves of wavelength \( \lambda \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A stretched string fixed at both ends can support standing waves. The fixed ends are nodes (points of no displacement), and between them, there can be antinodes (points of maximum displacement).

2. **Identifying the Harmonics**:
- For the fundamental frequency (first harmonic, \( n=1 \)), the string will have one antinode in the middle and two nodes at the ends. The length of the string \( l \) corresponds to half the wavelength of the wave.
- For the first harmonic, the relationship is:
\[ l = \frac{\lambda}{2} \]

3. **Generalizing for Higher Harmonics**:
- For the second harmonic (\( n=2 \)), the string will have two antinodes and three nodes. The length of the string corresponds to the full wavelength:
\[ l = \lambda \]
- For the third harmonic (\( n=3 \)), there will be three antinodes and four nodes:
\[ l = \frac{3\lambda}{2} \]

4. **Establishing the General Formula**:
- In general, for the \( n \)-th harmonic, the relationship can be expressed as:
\[ l = \frac{n\lambda}{2} \]
- Rearranging this gives us the wavelength in terms of the length of the string and the harmonic number:
\[ \lambda = \frac{2l}{n} \]

5. **Conclusion**:
- The wavelength \( \lambda \) of the stationary waves on a stretched string of length \( l \) fixed at both ends is given by:
\[ \lambda = \frac{2l}{n} \]

### Final Answer:
The wavelength \( \lambda \) of the stationary waves is:
\[ \lambda = \frac{2l}{n} \]