A particle exectuing S.H.M has potential energy `E_(1)` and `E_(2)` for displacement `x_(1)` and `x_(2)` respectively . The potential energy at a displacement `X_(1)+x_(2)` is

Options

A. `E_(1)+E_(2)`
B. `sqrt(E_(1)E_(2))`
C. `E_(1)^(2)+E_(2)^(2)`
D. `E_(1)+E_(2)+2sqrt(E_(1)E_(2))`

Explanation

To find the potential energy of a particle executing Simple Harmonic Motion (SHM) at a displacement of \( x_1 + x_2 \), we can follow these steps:

### Step 1: Understand the formula for potential energy in SHM
The potential energy \( E \) of a particle in SHM at a displacement \( x \) is given by:
\[ E = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

### Step 2: Write the potential energy at displacements \( x_1 \) and \( x_2 \)
For displacement \( x_1 \):
\[ E_1 = \frac{1}{2} k x_1^2 \]
For displacement \( x_2 \):
\[ E_2 = \frac{1}{2} k x_2^2 \]

### Step 3: Express \( x_1 \) and \( x_2 \) in terms of \( E_1 \) and \( E_2 \)
From the equations above, we can express \( x_1 \) and \( x_2 \) as:
\[ x_1 = \sqrt{\frac{2E_1}{k}} \]
\[ x_2 = \sqrt{\frac{2E_2}{k}} \]

### Step 4: Calculate the potential energy at displacement \( x_1 + x_2 \)
Now, we need to find the potential energy at the displacement \( x_1 + x_2 \):
\[ E = \frac{1}{2} k (x_1 + x_2)^2 \]
Expanding this, we get:
\[ E = \frac{1}{2} k (x_1^2 + 2x_1x_2 + x_2^2) \]

### Step 5: Substitute \( x_1^2 \) and \( x_2^2 \)
Substituting \( x_1^2 \) and \( x_2^2 \) from Step 2:
\[ E = \frac{1}{2} k \left( \frac{2E_1}{k} + 2x_1x_2 + \frac{2E_2}{k} \right) \]
This simplifies to:
\[ E = E_1 + E_2 + k x_1 x_2 \]

### Step 6: Substitute \( x_1 \) and \( x_2 \)
Now, substituting \( x_1 \) and \( x_2 \):
\[ E = E_1 + E_2 + k \left( \sqrt{\frac{2E_1}{k}} \cdot \sqrt{\frac{2E_2}{k}} \right) \]
This further simplifies to:
\[ E = E_1 + E_2 + 2 \sqrt{E_1 E_2} \]

### Final Result
Thus, the potential energy at the displacement \( x_1 + x_2 \) is:
\[ E = E_1 + E_2 + 2 \sqrt{E_1 E_2} \]