A particle executing SHM with an amplitude A. The displacement of the particle when its potential energy is half of its total energy is

Options

A. `(A)/(sqrt(2))`
B. `(A)/(2)`
C. `(A)/(4)`
D. `(A)/(3)`

Explanation

To solve the problem step by step, we need to analyze the relationship between potential energy, total energy, and displacement in simple harmonic motion (SHM).

### Step 1: Understand the Energy in SHM
In SHM, the total mechanical energy (E) of the system is constant and is given by the formula:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude of the motion.

### Step 2: Write the Expression for Potential Energy
The potential energy (PE) of a particle in SHM at a displacement \( x \) from the mean position is given by:
\[ PE = \frac{1}{2} k x^2 \]

### Step 3: Set Up the Equation for the Given Condition
According to the problem, we need to find the displacement \( x \) when the potential energy is half of the total energy. This can be expressed as:
\[ PE = \frac{1}{2} E \]
Substituting the expressions for PE and E, we have:
\[ \frac{1}{2} k x^2 = \frac{1}{2} \left( \frac{1}{2} k A^2 \right) \]

### Step 4: Simplify the Equation
Cancelling \( \frac{1}{2} \) from both sides gives:
\[ k x^2 = \frac{1}{2} k A^2 \]
Now, we can cancel \( k \) (assuming \( k \neq 0 \)):
\[ x^2 = \frac{1}{2} A^2 \]

### Step 5: Solve for Displacement \( x \)
Taking the square root of both sides, we find:
\[ x = \sqrt{\frac{1}{2}} A = \frac{A}{\sqrt{2}} \]

### Conclusion
Thus, the displacement of the particle when its potential energy is half of its total energy is:
\[ x = \frac{A}{\sqrt{2}} \]