A block is hanged from spring is a cage. Elogation is spring is `x_(1)=4sqrt2 nm` and `x_(2)=3sqrt2mm` mm when cage moves up and down respectively with same acceleration. The expansion (in mm) in spring when the cage moves horizontally with the same acceleration .

Answer

To solve the problem step by step, we will analyze the situation of the block hanging from the spring in the cage and how the elongation of the spring changes with different movements of the cage.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A block is hanging from a spring in a cage.
- The elongation of the spring when the cage moves up with acceleration \( a \) is given as \( x_1 = 4\sqrt{2} \, \text{nm} \).
- The elongation of the spring when the cage moves down with the same acceleration is \( x_2 = 3\sqrt{2} \, \text{mm} \).
- We need to find the elongation of the spring when the cage moves horizontally with the same acceleration.

2. **Effective Acceleration**:
- When the cage moves upward, the effective acceleration acting on the block is \( g + a \).
- When the cage moves downward, the effective acceleration is \( g - a \).

3. **Using Hooke's Law**:
- According to Hooke's law, the force exerted by the spring is proportional to its elongation:
\[ F = kx \]
- Where \( k \) is the spring constant and \( x \) is the elongation.

4. **Setting Up the Equations**:
- For upward motion:
\[ kx_1 = m(g + a) \quad \text{(1)} \]
- For downward motion:
\[ kx_2 = m(g - a) \quad \text{(2)} \]

5. **Substituting the Given Values**:
- From equation (1):
\[ k(4\sqrt{2} \, \text{nm}) = m(g + a) \]
- From equation (2):
\[ k(3\sqrt{2} \, \text{mm}) = m(g - a) \]

6. **Equating the Two Expressions**:
- Rearranging both equations gives:
\[ \frac{g + a}{4\sqrt{2}} = \frac{g - a}{3\sqrt{2}} \]
- Cross-multiplying yields:
\[ 3\sqrt{2}(g + a) = 4\sqrt{2}(g - a) \]
- Simplifying this gives:
\[ 3g + 3a = 4g - 4a \]
- Rearranging terms:
\[ 7a = g \quad \Rightarrow \quad a = \frac{g}{7} \quad \text{(3)} \]

7. **Finding the Elongation for Horizontal Motion**:
- When the cage moves horizontally with acceleration \( a \), the effective acceleration is:
\[ g_{\text{eff}} = \sqrt{a^2 + g^2} \]
- Substituting \( a \) from equation (3):
\[ g_{\text{eff}} = \sqrt{\left(\frac{g}{7}\right)^2 + g^2} = \sqrt{\frac{g^2}{49} + g^2} \]
- Combining the terms:
\[ g_{\text{eff}} = \sqrt{\frac{g^2 + 49g^2}{49}} = \sqrt{\frac{50g^2}{49}} = \frac{g\sqrt{50}}{7} \]

8. **Using Hooke's Law Again**:
- The elongation \( x \) for horizontal motion is given by:
\[ kx = mg_{\text{eff}} \quad \Rightarrow \quad x = \frac{mg_{\text{eff}}}{k} \]
- Substituting \( g_{\text{eff}} \):
\[ x = \frac{m \cdot \frac{g\sqrt{50}}{7}}{k} \]

9. **Relating to Previous Cases**:
- From previous elongation equations, we can find the relationship and solve for \( x \):
\[ x = 5 \, \text{mm} \]

### Final Answer:
The expansion in the spring when the cage moves horizontally with the same acceleration is \( x = 5 \, \text{mm} \).