The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal when displacement (amplitude = a) is

Options

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

Explanation

To find the displacement \( x \) at which the kinetic energy (KE) and potential energy (PE) of a particle executing simple harmonic motion (SHM) are equal, we can follow these steps:

### Step 1: Write the expressions for kinetic energy and potential energy in SHM.
- The potential energy (PE) of a particle in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} k (a^2 - x^2) \]
where \( a \) is the amplitude of the motion.

### Step 2: Set the kinetic energy equal to the potential energy.
According to the problem, we need to find the displacement \( x \) when \( KE = PE \):
\[ \frac{1}{2} k x^2 = \frac{1}{2} k (a^2 - x^2) \]

### Step 3: Simplify the equation.
We can cancel \( \frac{1}{2} k \) from both sides since \( k \) is a constant and not equal to zero:
\[ x^2 = a^2 - x^2 \]

### Step 4: Rearrange the equation.
Now, add \( x^2 \) to both sides:
\[ x^2 + x^2 = a^2 \]
This simplifies to:
\[ 2x^2 = a^2 \]

### Step 5: Solve for \( x^2 \).
Now, divide both sides by 2:
\[ x^2 = \frac{a^2}{2} \]

### Step 6: Take the square root to find \( x \).
Taking the square root of both sides gives:
\[ x = \frac{a}{\sqrt{2}} \]

### Conclusion:
Thus, the displacement \( x \) at which the kinetic energy and potential energy are equal is:
\[ x = \frac{a}{\sqrt{2}} \]