If `E_(P)andE_(K)` represent the potential energy and kinetic energy of a body undergoing S.H.M, then (E is the total energy of the body), at a position where the displacement is half the amplitude.

Options

A. `E_(P)=(E)/(2),E_(K)=(E)/(2)`
B. `E_(P)=(3E)/(4),E_(K)=(E)/(4)`
C. `E_(P)=(E)/(4),E_(K)=(3E)/(4)`
D. `E_(P)=(E)/(3),E_(K)=(2E)/(3)`

Explanation

To solve the problem of determining the potential energy \( E_P \) and kinetic energy \( E_K \) of a body undergoing Simple Harmonic Motion (SHM) at a position where the displacement is half the amplitude, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( A \) = Amplitude of the SHM
- \( x \) = Displacement from the equilibrium position
- \( E \) = Total energy of the system
- \( E_P \) = Potential energy
- \( E_K \) = Kinetic energy

Given that the displacement \( x = \frac{A}{2} \).

### Step 2: Write the Expressions for Kinetic and Potential Energy
The total energy \( E \) in SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( m \) is the mass of the body and \( \omega \) is the angular frequency.

The potential energy \( E_P \) at a displacement \( x \) is given by:
\[ E_P = \frac{1}{2} m \omega^2 x^2 \]
The kinetic energy \( E_K \) can be expressed as:
\[ E_K = E - E_P \]

### Step 3: Calculate Potential Energy at \( x = \frac{A}{2} \)
Substituting \( x = \frac{A}{2} \) into the potential energy formula:
\[ E_P = \frac{1}{2} m \omega^2 \left(\frac{A}{2}\right)^2 = \frac{1}{2} m \omega^2 \frac{A^2}{4} = \frac{1}{8} m \omega^2 A^2 \]

### Step 4: Relate Potential Energy to Total Energy
Now, we can express \( E_P \) in terms of total energy \( E \):
\[ E_P = \frac{1}{8} m \omega^2 A^2 = \frac{1}{4} \left(\frac{1}{2} m \omega^2 A^2\right) = \frac{1}{4} E \]

### Step 5: Calculate Kinetic Energy
Using the relationship \( E_K = E - E_P \):
\[ E_K = E - E_P = E - \frac{1}{4} E = \frac{3}{4} E \]

### Conclusion
Thus, at a position where the displacement is half the amplitude:
- The potential energy \( E_P \) is \( \frac{1}{4} E \)
- The kinetic energy \( E_K \) is \( \frac{3}{4} E \)

### Final Answer
- \( E_P = \frac{1}{4} E \)
- \( E_K = \frac{3}{4} E \)