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. `sqrt2 cm`
C. 3 cm
D. `2sqrt2 cm`

Answer

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

### Step-by-Step Solution:

1. **Understand the Total Energy in SHM**:
The total mechanical energy (E) in simple harmonic motion is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude.

2. **Define Potential and Kinetic Energy**:
The potential energy (PE) at a displacement \( x \) is given by:
\[ PE = \frac{1}{2} k x^2 \]
The kinetic energy (KE) at the same displacement is given by:
\[ KE = E - PE \]

3. **Set Up the Condition for Half Energy**:
We want the potential energy to be equal to the kinetic energy:
\[ PE = KE \]
Since \( KE = E - PE \), we can write:
\[ PE = \frac{1}{2} E \]
Therefore, we have:
\[ \frac{1}{2} k x^2 = \frac{1}{2} E \]

4. **Substituting Total Energy**:
Substitute \( E \) into the equation:
\[ \frac{1}{2} k x^2 = \frac{1}{2} \left(\frac{1}{2} k A^2\right) \]
This simplifies to:
\[ k x^2 = \frac{1}{2} k A^2 \]

5. **Canceling \( k \)**:
Assuming \( k \neq 0 \), we can divide both sides by \( k \):
\[ x^2 = \frac{1}{2} A^2 \]

6. **Finding Displacement \( x \)**:
Taking the square root of both sides gives:
\[ x = \pm \frac{A}{\sqrt{2}} \]
Given that the amplitude \( A = 4 \, \text{cm} \):
\[ x = \pm \frac{4}{\sqrt{2}} = \pm 2\sqrt{2} \, \text{cm} \]

7. **Final Result**:
Thus, the displacement from the equilibrium position where the energy is half potential and half kinetic is:
\[ x \approx \pm 2.83 \, \text{cm} \]