Two springs A and B`(k_A=2k_B)` are stretched by applying forces of equal magnitudes at the force 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 find the energy stored in spring B when the energy stored in spring A is given as E, and we know the relationship between the spring constants of the two springs.

### Step-by-step Solution:

1. **Understanding Spring Constants**:
We know that the spring constants are related as \( k_A = 2k_B \). This means that spring A is stiffer than spring B.

2. **Force Applied**:
Both springs are stretched by applying forces of equal magnitude \( F \). According to Hooke's Law, the force exerted by a spring is given by:
\[ F = kx \]
where \( k \) is the spring constant and \( x \) is the elongation of the spring.

3. **Elongation of Each Spring**:
For spring A:
\[ F = k_A x_A \implies x_A = \frac{F}{k_A} \]
For spring B:
\[ F = k_B x_B \implies x_B = \frac{F}{k_B} \]

4. **Energy Stored in Each Spring**:
The potential energy stored in a spring is given by:
\[ U = \frac{1}{2} k x^2 \]
For spring A:
\[ U_A = \frac{1}{2} k_A x_A^2 = \frac{1}{2} k_A \left(\frac{F}{k_A}\right)^2 = \frac{F^2}{2k_A} \]
For spring B:
\[ U_B = \frac{1}{2} k_B x_B^2 = \frac{1}{2} k_B \left(\frac{F}{k_B}\right)^2 = \frac{F^2}{2k_B} \]

5. **Relating the Energies**:
Since \( k_A = 2k_B \), we can express \( U_A \) in terms of \( U_B \):
\[ U_A = \frac{F^2}{2k_A} = \frac{F^2}{2(2k_B)} = \frac{F^2}{4k_B} \]
Now substituting for \( U_B \):
\[ U_B = \frac{F^2}{2k_B} \]

6. **Finding the Ratio of Energies**:
Now, we can find the ratio of the energies:
\[ \frac{U_A}{U_B} = \frac{\frac{F^2}{4k_B}}{\frac{F^2}{2k_B}} = \frac{1/4}{1/2} = \frac{1}{2} \]
Thus, we have:
\[ U_B = 2U_A \]

7. **Final Result**:
If the energy stored in spring A is \( E \), then:
\[ U_B = 2E \]

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