Passage I) In simple harmonic motion force acting on a particle is given as `F=-4x`, total mechanical energy of the particle is 10 J and amplitude of oscillations is 2m, At time t=0 acceleration of the particle is `-16m/s^(2)`. Mass of the particle is 0.5 kg.
At x=+1m, potential energy and kinetic energy of the particle are

Options

A. 2 J and 8 J
B. 8 J and 2 J
C. 6 J and 4 J
D. 4 J and 6 J

Answer

To find the potential energy (PE) and kinetic energy (KE) of the particle at \( x = +1 \, \text{m} \), we can follow these steps:

### Step 1: Identify the Force Constant \( K \)
The force acting on the particle is given by:
\[ F = -4x \]
From this, we can identify the force constant \( K \):
\[ K = 4 \, \text{N/m} \]

### Step 2: Calculate the Angular Frequency \( \omega \)
The angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{K}{m}} \]
where \( m = 0.5 \, \text{kg} \). Substituting the values:
\[ \omega = \sqrt{\frac{4}{0.5}} = \sqrt{8} = 2\sqrt{2} \, \text{rad/s} \]

### Step 3: Calculate the Velocity \( v \) at \( x = 1 \, \text{m} \)
In simple harmonic motion, the velocity \( v \) at position \( x \) can be calculated using:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A = 2 \, \text{m} \) (the amplitude). Substituting the values:
\[ v = 2\sqrt{2} \sqrt{2^2 - 1^2} = 2\sqrt{2} \sqrt{4 - 1} = 2\sqrt{2} \sqrt{3} = 2\sqrt{6} \, \text{m/s} \]

### Step 4: Calculate the Kinetic Energy \( KE \)
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the mass and the calculated velocity:
\[ KE = \frac{1}{2} \times 0.5 \times (2\sqrt{6})^2 = \frac{1}{2} \times 0.5 \times 24 = 6 \, \text{J} \]

### Step 5: Calculate the Potential Energy \( PE \)
The total mechanical energy \( E \) is given as:
\[ E = KE + PE \]
We know \( E = 10 \, \text{J} \) and \( KE = 6 \, \text{J} \), so we can find \( PE \):
\[ PE = E - KE = 10 - 6 = 4 \, \text{J} \]

### Final Answer
At \( x = +1 \, \text{m} \):
- Potential Energy \( PE = 4 \, \text{J} \)
- Kinetic Energy \( KE = 6 \, \text{J} \)

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