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 need to analyze the elongation of the spring under different conditions of motion.

### Step 1: Understand the Forces Acting on the Block
When the block is hanging from the spring, the forces acting on it are:
- The weight of the block, \( mg \) (acting downwards).
- The spring force, \( kx \) (acting upwards).

### Step 2: Analyze the Case When the Cage Moves Upwards
When the cage moves upwards with acceleration \( a \):
- The effective force acting on the block is \( mg + ma \) (downwards).
- The spring force is \( kx_1 \) (upwards).

From this, we can write the equation:
\[ mg + ma = kx_1 \]
Rearranging gives:
\[ k = \frac{m(g + a)}{x_1} \]

### Step 3: Analyze the Case When the Cage Moves Downwards
When the cage moves downwards with the same acceleration \( a \):
- The effective force acting on the block is \( mg - ma \) (downwards).
- The spring force is \( kx_2 \) (upwards).

From this, we can write the equation:
\[ mg - ma = kx_2 \]
Rearranging gives:
\[ k = \frac{m(g - a)}{x_2} \]

### Step 4: Equate the Two Expressions for Spring Constant \( k \)
Since both expressions represent the same spring constant \( k \), we can set them equal to each other:
\[ \frac{m(g + a)}{x_1} = \frac{m(g - a)}{x_2} \]

### Step 5: Simplify the Equation
Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{g + a}{x_1} = \frac{g - a}{x_2} \]
Cross-multiplying gives:
\[ (g + a)x_2 = (g - a)x_1 \]

### Step 6: Solve for \( g \) and \( a \)
Expanding the equation:
\[ gx_2 + ax_2 = gx_1 - ax_1 \]
Rearranging gives:
\[ g(x_1 - x_2) = a(x_2 + x_1) \]
Thus, we can express \( a \) as:
\[ a = \frac{g(x_1 - x_2)}{x_1 + x_2} \]

### Step 7: Analyze the Case When the Cage Moves Horizontally
When the cage moves horizontally with the same acceleration \( a \):
- The vertical forces remain the same: \( mg \) (downwards) and \( kx \) (upwards).
- The horizontal acceleration creates a pseudo force \( ma \) acting horizontally.

The resultant elongation \( x \) in the spring can be calculated using Pythagorean theorem:
\[ kx = \sqrt{(mg)^2 + (ma)^2} \]
Thus,
\[ x = \frac{1}{k} \sqrt{(mg)^2 + (ma)^2} \]

### Step 8: Substitute for \( a \)
Substituting \( a \) from the earlier step:
\[ x = \frac{1}{k} \sqrt{(mg)^2 + \left(\frac{g(x_1 - x_2)}{x_1 + x_2}\right)^2 m^2} \]

### Step 9: Final Expression for Elongation
Using the expressions for \( k \) derived earlier, we can substitute and simplify to find the final elongation \( x \).

### Final Answer
The elongation in the spring when the cage moves horizontally with the same acceleration is given by:
\[ x = \frac{1}{k} \sqrt{(mg)^2 + \left(\frac{g(x_1 - x_2)}{x_1 + x_2}\right)^2 m^2} \]