One end of a taut string of length `3m` along the x-axis is fixed at `x = 0`. The speed of the waves in the string is `100ms^(-1)`. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible wavelength`(s)` of these sationary waves is (are)

Options

A. `y(t)=Asin(2pix)/6cos(50pit)/3`
B. `y(t)=Asin(pix)/3cos(100pit)/3`
C. `y(t)=Asin(5pix)/6cos(255pit)/3`
D. `y(t)=Asin(5pix)/2cos250pit`

Explanation

To solve the problem, we need to determine the possible wavelengths of stationary waves on a taut string of length 3 m, given that the speed of the waves in the string is 100 m/s.

### Step-by-Step Solution:

1. **Understanding the Problem**:
The string is fixed at one end (x = 0) and vibrating at the other end. This setup allows for stationary waves to form. The equation for stationary waves can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t) \]
where \( k \) is the wave number and \( \omega \) is the angular frequency.

2. **Wave Speed Relation**:
The speed of the wave \( v \) is related to \( \omega \) and \( k \) by the equation:
\[ v = \frac{\omega}{k} \]
Given that \( v = 100 \, \text{m/s} \), we can use this relationship to find the possible values of \( k \) and \( \omega \).

3. **Boundary Conditions**:
For stationary waves on a string fixed at one end, the wavelengths must satisfy the condition:
\[ L = \frac{n\lambda}{2} \]
where \( L \) is the length of the string (3 m), \( n \) is a positive integer (1, 2, 3,...), and \( \lambda \) is the wavelength. Rearranging gives:
\[ \lambda = \frac{2L}{n} = \frac{2 \times 3}{n} = \frac{6}{n} \]

4. **Calculating Possible Wavelengths**:
Now we can calculate the possible wavelengths for different values of \( n \):
- For \( n = 1 \):
\[ \lambda_1 = \frac{6}{1} = 6 \, \text{m} \]
- For \( n = 2 \):
\[ \lambda_2 = \frac{6}{2} = 3 \, \text{m} \]
- For \( n = 3 \):
\[ \lambda_3 = \frac{6}{3} = 2 \, \text{m} \]
- For \( n = 4 \):
\[ \lambda_4 = \frac{6}{4} = 1.5 \, \text{m} \]
- For \( n = 5 \):
\[ \lambda_5 = \frac{6}{5} = 1.2 \, \text{m} \]
- Continuing this process gives us wavelengths for higher integers.

5. **Conclusion**:
The possible wavelengths of stationary waves in the string are:
\[ \lambda = \frac{6}{n} \text{ for } n = 1, 2, 3, 4, 5, \ldots \]
Therefore, the possible wavelengths are \( 6 \, \text{m}, 3 \, \text{m}, 2 \, \text{m}, 1.5 \, \text{m}, 1.2 \, \text{m}, \ldots \)