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 regarding the simple harmonic oscillator (SHO) where the potential energy (PE) is equal to the kinetic energy (KE), we can follow these steps:

### Step 1: Write the expressions for potential energy and kinetic energy
The potential energy (PE) of a simple harmonic oscillator is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the mean position.

The kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the oscillator and \( v \) is its velocity.

### Step 2: Relate velocity to displacement
In simple harmonic motion, the velocity can be expressed in terms of displacement using the equation:
\[ v^2 = \omega^2 a^2 - \omega^2 x^2 \]
where \( \omega \) is the angular frequency and \( a \) is the amplitude of the motion. This can be rearranged to:
\[ v^2 = \omega^2 (a^2 - x^2) \]

### Step 3: Substitute the expression for velocity into the kinetic energy formula
Substituting the expression for \( v^2 \) into the kinetic energy formula, we get:
\[ KE = \frac{1}{2} m (\omega^2 (a^2 - x^2)) \]
This simplifies to:
\[ KE = \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 x^2 \]

### Step 4: Use the relation between \( k \), \( m \), and \( \omega \)
The spring constant \( k \) is related to mass \( m \) and angular frequency \( \omega \) by:
\[ k = m \omega^2 \]
Substituting this into the potential energy expression, we have:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

### Step 5: Set potential energy equal to kinetic energy
To find the condition when potential energy equals kinetic energy, we set:
\[ \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} k x^2 \]
Substituting \( k = m \omega^2 \):
\[ \frac{1}{2} m \omega^2 a^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 x^2 \]

### Step 6: Simplify the equation
Cancelling \( \frac{1}{2} m \omega^2 \) from both sides gives:
\[ a^2 - x^2 = x^2 \]
This simplifies to:
\[ a^2 = 2x^2 \]

### Step 7: Solve for \( x \)
Taking the square root of both sides, we find:
\[ x = \frac{a}{\sqrt{2}} \]

### Step 8: Determine how many times the particle passes through this position in one cycle
In one complete cycle of motion, the particle will pass through the position \( x = \frac{a}{\sqrt{2}} \) twice (once while moving towards the maximum amplitude and once while returning).

### Conclusion
Thus, the answer is that the particle passes through the position where potential energy equals kinetic energy **4 times** in one complete cycle (twice for positive and twice for negative displacement).