For simple Harmonic Oscillator, the potential energy is equal to kinetic energy

Options

A. twice during each cycle
B. four times during each cycle
C. when x = 0
D. whe x = a

Answer

To solve the problem of when the potential energy is equal to the kinetic energy in a simple harmonic oscillator, we can follow these steps:

### Step 1: Understand the Energy Equations
In a simple harmonic oscillator, the total mechanical energy (E) is the sum of kinetic energy (KE) and potential energy (PE). The formulas for these energies are:
- Kinetic Energy (KE) = \( \frac{1}{2} m v^2 \)
- Potential Energy (PE) = \( \frac{1}{2} k x^2 \)

Where:
- \( m \) = mass of the oscillator
- \( v \) = velocity of the oscillator
- \( k \) = spring constant
- \( x \) = displacement from the mean position

### Step 2: Set Kinetic Energy Equal to Potential Energy
We are given that the potential energy is equal to the kinetic energy:
\[ PE = KE \]

Substituting the equations for PE and KE, we have:
\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]

### Step 3: Use the Relationship Between Velocity and Displacement
In simple harmonic motion, the velocity can be expressed in terms of displacement:
\[ v = \sqrt{\omega^2 (A^2 - x^2)} \]
Where:
- \( \omega \) = angular frequency
- \( A \) = amplitude of the motion

### Step 4: Substitute Velocity into the Energy Equation
Substituting for \( v \) in the kinetic energy equation gives:
\[ \frac{1}{2} k x^2 = \frac{1}{2} m \left( \omega^2 (A^2 - x^2) \right) \]

### Step 5: Simplify the Equation
Cancel the \( \frac{1}{2} \) from both sides:
\[ k x^2 = m \omega^2 (A^2 - x^2) \]

### Step 6: Use the Relationship Between \( k \), \( m \), and \( \omega \)
Recall that \( \omega^2 = \frac{k}{m} \). Substituting this into the equation gives:
\[ k x^2 = k (A^2 - x^2) \]

### Step 7: Rearrange and Solve for \( x \)
Rearranging the equation:
\[ k x^2 + k x^2 = k A^2 \]
\[ 2k x^2 = k A^2 \]
Dividing both sides by \( k \):
\[ 2x^2 = A^2 \]
\[ x^2 = \frac{A^2}{2} \]
Taking the square root:
\[ x = \frac{A}{\sqrt{2}} \]

### Step 8: Conclusion
Thus, the displacement at which the potential energy is equal to the kinetic energy is \( x = \frac{A}{\sqrt{2}} \).

### Step 9: Determine How Many Times This Occurs in One Cycle
In one complete cycle of simple harmonic motion, the particle will cross the position where potential energy equals kinetic energy four times (twice while moving towards the equilibrium position and twice while moving away).

### Final Answer
The potential energy is equal to the kinetic energy at \( x = \frac{A}{\sqrt{2}} \), and this occurs four times in one complete oscillation.

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