The kinetic energy and the potential energy of a particle executing `SHM` are equal The ratio of its displacement and amplitude will be

Options

A. `(1)/(sqrt(2))`
B. `sqrt(3)/(2)`
C. `(1)/(2)`
D. `sqrt(2)`

Explanation

To solve the problem of finding the ratio of displacement (x) to amplitude (a) when the kinetic energy (KE) and potential energy (PE) of a particle executing Simple Harmonic Motion (SHM) are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy in SHM**:
- The total mechanical energy (E) in SHM is given by the formula:
\[ E = \frac{1}{2} k a^2 \]
- Here, \(k\) is the spring constant and \(a\) is the amplitude.

2. **Potential Energy (PE)**:
- The potential energy when the particle is at displacement \(x\) is given by:
\[ PE = \frac{1}{2} k x^2 \]

3. **Kinetic Energy (KE)**:
- The kinetic energy can be derived from the total energy:
\[ KE = E - PE = \frac{1}{2} k a^2 - \frac{1}{2} k x^2 \]

4. **Set KE Equal to PE**:
- According to the problem, KE = PE. Therefore, we can write:
\[ \frac{1}{2} k a^2 - \frac{1}{2} k x^2 = \frac{1}{2} k x^2 \]

5. **Simplify the Equation**:
- Rearranging the equation gives:
\[ \frac{1}{2} k a^2 = k x^2 \]
- Dividing both sides by \(\frac{1}{2} k\) (assuming \(k \neq 0\)):
\[ a^2 = 2 x^2 \]

6. **Find the Ratio**:
- Taking the square root of both sides:
\[ a = \sqrt{2} x \]
- Rearranging gives:
\[ \frac{x}{a} = \frac{1}{\sqrt{2}} \]

7. **Conclusion**:
- Thus, the ratio of displacement to amplitude is:
\[ \frac{x}{a} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The ratio of displacement to amplitude is \(\frac{1}{\sqrt{2}}\).

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