The stationary waves set up on a string have the equation :
`y = ( 2 mm) sin [ (6.28 m^(-1)) x] cos omega t`
The stationary wave is created by two identical waves , of amplitude `A` each , moving in opposite directions along the string . Then :

Options

A. A = 2 mm
B. A = 1 mm
C. The smallest length of the string is 50 cm
D. The smallest length of the string is 2 m

Explanation

To solve the problem, we will analyze the given equation of the stationary wave and compare it to the standard form of a stationary wave equation.

### Step-by-step Solution:

1. **Identify the given wave equation**:
The equation of the stationary wave is given as:
\[ y = (2 \, \text{mm}) \sin(6.28 \, \text{m}^{-1} \, x) \cos(\omega t) \]

2. **Compare with the standard form**:
The standard form of a stationary wave can be expressed as:
\[ y = 2A \sin\left(\frac{n\pi}{L} x\right) \cos(\omega t) \]
where \( A \) is the amplitude, \( n \) is the mode number, and \( L \) is the length of the string.

3. **Extract the amplitude**:
From the equation, we see that:
\[ 2A = 2 \, \text{mm} \]
Therefore, the amplitude \( A \) is:
\[ A = 1 \, \text{mm} \]

4. **Identify the wave number**:
We have:
\[ \frac{n\pi}{L} = 6.28 \, \text{m}^{-1} \]
We know that \( 6.28 \, \text{m}^{-1} \) can be rewritten as \( 2\pi \, \text{m}^{-1} \). Thus, we can equate:
\[ \frac{n\pi}{L} = 2\pi \]

5. **Cancel out \(\pi\)**:
Dividing both sides by \(\pi\):
\[ \frac{n}{L} = 2 \]

6. **Solve for the length \( L \)**:
Rearranging gives:
\[ L = \frac{n}{2} \]
The smallest value for \( n \) (the fundamental mode) is 1. Therefore:
\[ L = \frac{1}{2} = 0.5 \, \text{m} \quad \text{or} \quad 50 \, \text{cm} \]

7. **Conclusion**:
The amplitude of the wave is \( 1 \, \text{mm} \) and the smallest length of the string is \( 50 \, \text{cm} \).

### Final Answers:
- Amplitude \( A = 1 \, \text{mm} \)
- Smallest length \( L = 50 \, \text{cm} \)