A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm . At what displacement from the equilibrium position, is its energy half potential and half kinetic

Options

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

Explanation

To solve the problem of finding the displacement from the equilibrium position where the energy of a particle in simple harmonic motion is half potential and half kinetic, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We know that the total mechanical energy (E) in simple harmonic motion (SHM) is the sum of kinetic energy (KE) and potential energy (PE). We need to find the displacement (x) from the equilibrium position where KE = PE.

2. **Total Energy in SHM**: The total energy (E) in SHM is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \(A\) is the amplitude.

3. **Potential Energy in SHM**: The potential energy (PE) at displacement \(x\) is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

4. **Kinetic Energy in SHM**: The kinetic energy (KE) at displacement \(x\) is given by:
\[ KE = E - PE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 \]

5. **Setting KE = PE**: Since we want to find the position where KE = PE, we can set:
\[ KE = PE \]
This implies:
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 \]

6. **Simplifying the Equation**: Rearranging gives:
\[ x^2 = A^2 - x^2 \]
Therefore,
\[ 2x^2 = A^2 \]

7. **Finding x**: From the equation \(2x^2 = A^2\), we can solve for \(x\):
\[ x^2 = \frac{A^2}{2} \]
Taking the square root gives:
\[ x = \pm \frac{A}{\sqrt{2}} \]

8. **Substituting the Amplitude**: Given that the amplitude \(A = 4 \text{ cm}\):
\[ x = \pm \frac{4}{\sqrt{2}} = \pm 2\sqrt{2} \text{ cm} \]

9. **Conclusion**: The displacement from the equilibrium position where the energy is half potential and half kinetic is:
\[ x = \pm 2\sqrt{2} \text{ cm} \]