The displacement of a particle of mass `3g` executing simple harmonic motion is given by `x =3sin(0.2t)` in `SI` units. The kinetic energy of the particle at a point which is at a displacement equal to `1/3` of its amplitude from its mean position is

Options

A. `12 xx 10^(-3)` J
B. `25 xx 10^(-3)` J
C. `0.48 xx 10^(-3)` J
D. `0.24 xx 10^(-3)` J

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 3 \, \text{g} = 3 \times 10^{-3} \, \text{kg} \)
- Displacement equation: \( x = 3 \sin(0.2t) \)

### Step 2: Determine the amplitude and angular frequency
From the displacement equation \( x = A \sin(\omega t) \):
- Amplitude \( A = 3 \, \text{m} \)
- Angular frequency \( \omega = 0.2 \, \text{rad/s} \)

### Step 3: Calculate the displacement at \( \frac{1}{3} \) of the amplitude
- Displacement \( x = \frac{1}{3} A = \frac{1}{3} \times 3 = 1 \, \text{m} \)

### Step 4: Use the kinetic energy formula for SHM
The kinetic energy \( KE \) of a particle in simple harmonic motion can be calculated using the formula:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
Substituting the known values:
- \( m = 3 \times 10^{-3} \, \text{kg} \)
- \( \omega = 0.2 \, \text{rad/s} \)
- \( A = 3 \, \text{m} \)
- \( x = 1 \, \text{m} \)

### Step 5: Substitute the values into the kinetic energy formula
\[ KE = \frac{1}{2} (3 \times 10^{-3}) (0.2)^2 \left(3^2 - 1^2\right) \]
Calculating \( A^2 - x^2 \):
\[ A^2 = 3^2 = 9, \quad x^2 = 1^2 = 1 \quad \Rightarrow \quad A^2 - x^2 = 9 - 1 = 8 \]

### Step 6: Calculate \( \omega^2 \)
\[ \omega^2 = (0.2)^2 = 0.04 \]

### Step 7: Substitute back into the kinetic energy equation
\[ KE = \frac{1}{2} (3 \times 10^{-3}) (0.04) (8) \]
Calculating:
\[ KE = \frac{1}{2} (3 \times 10^{-3}) (0.32) = \frac{1}{2} \times 0.00096 = 0.00048 \, \text{J} \]

### Step 8: Final result
\[ KE = 48 \times 10^{-5} \, \text{J} = 0.48 \times 10^{-3} \, \text{J} \]

### Conclusion
The kinetic energy of the particle at a displacement equal to \( \frac{1}{3} \) of its amplitude from its mean position is \( 0.48 \times 10^{-3} \, \text{J} \).

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