When the displacement of a particle executing SHM is one-fourth of its amplitude, what fraction of the total energy is the kinetic energy?

Options

A. `(16)/(15)`
B. `(15)/(16)`
C. `(3)/(4)`
D. `(4)/(3)`

Explanation

To solve the problem, we need to determine the fraction of kinetic energy (KE) when the displacement (x) of a particle executing Simple Harmonic Motion (SHM) is one-fourth of its amplitude (A).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Displacement \( x = \frac{A}{4} \)
- Amplitude \( A \)

2. **Formulas for Kinetic Energy and Total Energy in SHM**:
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The total energy (TE) in SHM is given by:
\[ TE = \frac{1}{2} m \omega^2 A^2 \]

3. **Finding the Fraction of Kinetic Energy to Total Energy**:
- We need to find the ratio \( \frac{KE}{TE} \):
\[ \frac{KE}{TE} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 A^2} \]
- The \( \frac{1}{2} m \omega^2 \) terms cancel out:
\[ \frac{KE}{TE} = \frac{A^2 - x^2}{A^2} \]

4. **Substituting the Value of Displacement**:
- Substitute \( x = \frac{A}{4} \):
\[ x^2 = \left(\frac{A}{4}\right)^2 = \frac{A^2}{16} \]
- Now substitute \( x^2 \) into the equation:
\[ \frac{KE}{TE} = \frac{A^2 - \frac{A^2}{16}}{A^2} \]

5. **Simplifying the Expression**:
- Simplifying the numerator:
\[ A^2 - \frac{A^2}{16} = \frac{16A^2}{16} - \frac{A^2}{16} = \frac{15A^2}{16} \]
- Now substitute this back into the ratio:
\[ \frac{KE}{TE} = \frac{\frac{15A^2}{16}}{A^2} \]
- The \( A^2 \) terms cancel out:
\[ \frac{KE}{TE} = \frac{15}{16} \]

6. **Conclusion**:
- Therefore, the fraction of the total energy that is kinetic energy when the displacement is one-fourth of the amplitude is:
\[ \frac{KE}{TE} = \frac{15}{16} \]

### Final Answer:
The fraction of the total energy that is kinetic energy is \( \frac{15}{16} \).