In SHM, kinetic energy is `(1//4)^(th)` of the total energy at a displacement equal to
(Here A is the amplitude of oscillations.)

Options

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

Explanation

To solve the problem, we need to find the displacement \( x \) at which the kinetic energy \( K \) of a particle in simple harmonic motion (SHM) is one-fourth of the total energy \( E \).

### Step-by-Step Solution:

1. **Understanding Total Energy in SHM**:
The total energy \( E \) of a particle in SHM is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( A \) is the amplitude, \( m \) is the mass of the particle, and \( \omega \) is the angular frequency.

2. **Kinetic Energy in SHM**:
The kinetic energy \( K \) of the particle at a displacement \( x \) is given by:
\[ K = \frac{1}{2} m v^2 \]
where \( v \) is the velocity of the particle. The velocity can be expressed in terms of \( x \) as:
\[ v = \omega \sqrt{A^2 - x^2} \]
Thus, substituting for \( v \):
\[ K = \frac{1}{2} m (\omega^2 (A^2 - x^2)) = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 \]

3. **Setting Up the Equation**:
According to the problem, the kinetic energy \( K \) is one-fourth of the total energy \( E \):
\[ K = \frac{1}{4} E \]
Substituting the expressions for \( K \) and \( E \):
\[ \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{4} \left(\frac{1}{2} m \omega^2 A^2\right) \]

4. **Simplifying the Equation**:
Simplifying the right side:
\[ \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{8} m \omega^2 A^2 \]
Now, multiplying through by \( 2 \) to eliminate the fraction:
\[ m \omega^2 A^2 - m \omega^2 x^2 = \frac{1}{4} m \omega^2 A^2 \]

5. **Rearranging Terms**:
Rearranging gives:
\[ m \omega^2 A^2 - \frac{1}{4} m \omega^2 A^2 = m \omega^2 x^2 \]
Simplifying the left side:
\[ \frac{3}{4} m \omega^2 A^2 = m \omega^2 x^2 \]

6. **Dividing by \( m \omega^2 \)**:
Dividing both sides by \( m \omega^2 \) (assuming \( m \) and \( \omega \) are not zero):
\[ \frac{3}{4} A^2 = x^2 \]

7. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ x = A \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} A \]

### Final Answer:
The displacement \( x \) at which the kinetic energy is one-fourth of the total energy is:
\[ x = \frac{\sqrt{3}}{2} A \]