If in a stationary wave the amplitude corresponding to antinode is 4 cm, then the amplitude corresponding to a particle of medium located exactly midway between a node and an antinode is :-

Options

A. 2cm
B. `2sqrt(2) cm `
C. `sqrt(2) cm `
D. 1.5 cm

Answer

To solve the problem, we need to determine the amplitude of a particle located exactly midway between a node and an antinode in a stationary wave, given that the amplitude at the antinode is 4 cm.

### Step-by-Step Solution:

1. **Understand the Positioning**:
- In a stationary wave, the points of maximum amplitude are called antinodes, and the points of zero amplitude are called nodes.
- The distance from a node to the nearest antinode is \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength.

2. **Identify the Midway Point**:
- The point exactly midway between a node and an antinode is located at a distance of \( \frac{\lambda}{8} \) from the node.

3. **Use the Wave Equation**:
- The amplitude \( A(x) \) at any point \( x \) in a stationary wave can be expressed as:
\[ A(x) = A \sin(kx) \]
- Here, \( A \) is the amplitude at the antinode, \( k \) is the wave number, and \( x \) is the position from the node.

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

5. **Substitute Values**:
- For our case, we need to find the amplitude at \( x = \frac{\lambda}{8} \):
\[ A\left(\frac{\lambda}{8}\right) = A \sin\left(k \cdot \frac{\lambda}{8}\right) \]
- Substituting \( k \):
\[ A\left(\frac{\lambda}{8}\right) = A \sin\left(\frac{2\pi}{\lambda} \cdot \frac{\lambda}{8}\right) = A \sin\left(\frac{\pi}{4}\right) \]

6. **Calculate the Amplitude**:
- We know that \( A = 4 \, \text{cm} \) (the amplitude at the antinode).
- The value of \( \sin\left(\frac{\pi}{4}\right) \) is \( \frac{1}{\sqrt{2}} \).
- Thus:
\[ A\left(\frac{\lambda}{8}\right) = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{cm} \]

### Final Answer:
The amplitude corresponding to a particle of the medium located exactly midway between a node and an antinode is \( 2\sqrt{2} \, \text{cm} \).