The equation of a standing wave in a string fixed at both its ends is given as y=2A sin kx cos `omegat`. The amplitude and frequency of a particle vibrating at the point of string midway between a node and an antinode is

Options

A. `A,(omega)/(2pi)`
B. `(A)/(sqrt(2)),(omega)/(2pi) `
C. `A, (omega)/(pi) `
D. `sqrt(2)A , (omega)/(2pi) `

Answer

To solve the problem, we need to find the amplitude and frequency of a particle vibrating at the point of the string midway between a node and an antinode, given the standing wave equation:

\[ y = 2A \sin(kx) \cos(\omega t) \]

### Step-by-Step Solution:

1. **Identify the Position Between Node and Antinode:**
- The distance between a node and an antinode is \( \frac{\lambda}{4} \).
- Therefore, the midpoint between a node and an antinode is at a distance of \( \frac{\lambda}{8} \) from the node.

2. **Substitute the Position into the Wave Equation:**
- We need to find \( y \) at \( x = \frac{\lambda}{8} \).
- Substitute \( x \) into the wave equation:
\[ y = 2A \sin\left(k \cdot \frac{\lambda}{8}\right) \cos(\omega t) \]

3. **Calculate \( k \):**
- The wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
- Substitute \( k \) into the equation:
\[ y = 2A \sin\left(\frac{2\pi}{\lambda} \cdot \frac{\lambda}{8}\right) \cos(\omega t) \]
- This simplifies to:
\[ y = 2A \sin\left(\frac{\pi}{4}\right) \cos(\omega t) \]

4. **Evaluate \( \sin\left(\frac{\pi}{4}\right) \):**
- The value of \( \sin\left(\frac{\pi}{4}\right) \) is:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
- Thus, substituting this value gives:
\[ y = 2A \cdot \frac{1}{\sqrt{2}} \cos(\omega t) = \sqrt{2}A \cos(\omega t) \]

5. **Determine the Amplitude:**
- The amplitude of the particle at this position is:
\[ \text{Amplitude} = \sqrt{2}A \]

6. **Determine the Frequency:**
- The frequency \( f \) is given by the formula:
\[ f = \frac{\omega}{2\pi} \]
- Since \( \omega \) is given in the equation, the frequency remains:
\[ \text{Frequency} = \frac{\omega}{2\pi} \]

### Final Answers:
- **Amplitude:** \( \sqrt{2}A \)
- **Frequency:** \( \frac{\omega}{2\pi} \)