Two springs A and B `(k_(A)=2k_(B))` are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is :

Options

A. `E//2`
B. 2E
C. E
D. `E//4`

Answer

To solve the problem, we need to analyze the energy stored in two springs A and B, given that the stiffness of spring A is twice that of spring B. Let's denote the stiffness of spring B as \( k_B \) and the stiffness of spring A as \( k_A = 2k_B \).

### Step-by-Step Solution:

1. **Identify the Forces in the Springs:**
Since equal forces are applied to both springs, we can denote the force applied to both springs as \( F \). Thus, we have:
\[ F = k_A x_A = k_B x_B \]
where \( x_A \) and \( x_B \) are the extensions of springs A and B, respectively.

2. **Relate the Extensions:**
Substituting \( k_A = 2k_B \) into the force equation gives:
\[ 2k_B x_A = k_B x_B \]
Dividing both sides by \( k_B \) (assuming \( k_B \neq 0 \)):
\[ 2x_A = x_B \]
This implies:
\[ x_B = 2x_A \]

3. **Calculate the Energy Stored in Spring A:**
The energy stored in a spring is given by the formula:
\[ E_A = \frac{1}{2} k_A x_A^2 \]
Substituting \( k_A = 2k_B \):
\[ E_A = \frac{1}{2} (2k_B) x_A^2 = k_B x_A^2 \]
We denote this energy as \( E \):
\[ E = k_B x_A^2 \]

4. **Calculate the Energy Stored in Spring B:**
Using the extension \( x_B = 2x_A \), we can find the energy stored in spring B:
\[ E_B = \frac{1}{2} k_B x_B^2 = \frac{1}{2} k_B (2x_A)^2 = \frac{1}{2} k_B (4x_A^2) = 2 k_B x_A^2 \]
Thus, we can express \( E_B \) in terms of \( E \):
\[ E_B = 2E \]

### Conclusion:
The energy stored in spring B is \( 2E \), where \( E \) is the energy stored in spring A.