In SHM for how many times potential energy is equal to kinetic energy during one complete period

Options

A. 1
B. 2
C. 4
D. 8

Explanation

To solve the question of how many times potential energy is equal to kinetic energy during one complete period of Simple Harmonic Motion (SHM), we can follow these steps:

### Step 1: Understand the Energy in SHM
In SHM, the total mechanical energy (E) is conserved and is the sum of kinetic energy (KE) and potential energy (PE). The expressions for kinetic and potential energy in SHM are:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = \( \frac{1}{2} k x^2 \)

Where:
- \( m \) is the mass of the body,
- \( v \) is the velocity,
- \( k \) is the spring constant,
- \( x \) is the displacement from the mean position.

### Step 2: Identify the Points of Equal Energy
The potential energy is maximum at the extreme positions (where \( x = \pm A \), A being the amplitude), and the kinetic energy is maximum at the mean position (where \( x = 0 \)).

At the extreme positions:
- KE = 0
- PE = maximum

At the mean position:
- KE = maximum
- PE = 0

### Step 3: Set PE Equal to KE
To find when PE equals KE, we set the two equations equal to each other:
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]
This simplifies to:
\[ m v^2 = k x^2 \]

### Step 4: Express Velocity in Terms of Displacement
Using the relationship \( v = \omega \sqrt{A^2 - x^2} \) (where \( \omega \) is the angular frequency), we can substitute for \( v \):
\[ m (\omega^2 (A^2 - x^2)) = k x^2 \]
Since \( k = m \omega^2 \), we can substitute \( k \):
\[ m \omega^2 (A^2 - x^2) = m \omega^2 x^2 \]
This simplifies to:
\[ A^2 - x^2 = x^2 \]
\[ A^2 = 2x^2 \]
\[ x^2 = \frac{A^2}{2} \]
Thus, the displacement \( x \) where PE equals KE is:
\[ x = \pm \frac{A}{\sqrt{2}} \]

### Step 5: Determine How Many Times This Occurs in One Period
In one complete cycle (or period), the body moves from one extreme position to the other and back. The points where \( x = \pm \frac{A}{\sqrt{2}} \) occur twice as the body moves towards the mean position and twice as it moves back to the other extreme position.

Thus, the total number of times that the potential energy is equal to the kinetic energy in one complete period is:
- Once while moving from the positive extreme to the mean position.
- Once while moving from the mean position to the negative extreme.
- Once while moving from the negative extreme back to the mean position.
- Once while moving from the mean position back to the positive extreme.

### Final Answer
Therefore, the potential energy is equal to the kinetic energy **four times** during one complete period of SHM.

---