The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude is

Options

A. `E//2`
B. `E//4`
C. `3 E//4`
D. `sqrt(3)//4 E`

Explanation

To solve the problem, we need to find the kinetic energy of a body executing Simple Harmonic Motion (S.H.M.) when its displacement is half of the amplitude. Let's break it down step by step.

### Step 1: Understand the Total Energy in S.H.M.
The total energy \( E \) of a body executing S.H.M. is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( m \) is the mass of the body,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion.

### Step 2: Determine the Displacement
We need to find the kinetic energy when the displacement \( x \) is half of the amplitude:
\[ x = \frac{A}{2} \]

### Step 3: Use the Velocity Formula in S.H.M.
The velocity \( v \) of a body in S.H.M. at a displacement \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting \( x = \frac{A}{2} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} \]
Calculating \( \left(\frac{A}{2}\right)^2 \):
\[ \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Thus, we have:
\[ v = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \omega \frac{A\sqrt{3}}{2} \]

### Step 4: Calculate the Kinetic Energy
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} mv^2 \]
Substituting the expression for \( v \):
\[ K = \frac{1}{2} m \left(\omega \frac{A\sqrt{3}}{2}\right)^2 \]
Calculating \( v^2 \):
\[ v^2 = \left(\omega \frac{A\sqrt{3}}{2}\right)^2 = \omega^2 \frac{3A^2}{4} \]
Thus, the kinetic energy becomes:
\[ K = \frac{1}{2} m \cdot \omega^2 \cdot \frac{3A^2}{4} = \frac{3}{8} m \omega^2 A^2 \]

### Step 5: Relate Kinetic Energy to Total Energy
From the total energy expression:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
We can express \( K \) in terms of \( E \):
\[ K = \frac{3}{8} m \omega^2 A^2 = \frac{3}{4} \cdot \frac{1}{2} m \omega^2 A^2 = \frac{3}{4} E \]

### Final Answer
Thus, the kinetic energy when the displacement is half of the amplitude is:
\[ K = \frac{3E}{4} \]