A particle of mass 40 g executes a simple harmonic motion of amplitude 2.0 cm. If the time period is 0.20 s, find the total mechanical energy of the system.

Options

A. `14.8xx10^(-3)J`
B. `7.9xx10^(-3)J`
C. `23.7xx10^(-3)J`
D. `3.9xx10^(-3)J`

Explanation

To find the total mechanical energy of a particle executing simple harmonic motion, we can follow these steps:

### Step 1: Convert the mass from grams to kilograms
Given mass \( m = 40 \, \text{g} \).

To convert grams to kilograms:
\[ m = 40 \, \text{g} = 40 \times 10^{-3} \, \text{kg} = 0.040 \, \text{kg} \]

### Step 2: Convert the amplitude from centimeters to meters
Given amplitude \( A = 2 \, \text{cm} \).

To convert centimeters to meters:
\[ A = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} = 0.02 \, \text{m} \]

### Step 3: Use the time period to find the angular frequency
Given time period \( T = 0.20 \, \text{s} \).

The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{0.20} = 10\pi \, \text{rad/s} \]

### Step 4: Calculate the total mechanical energy
The formula for total mechanical energy \( E \) in simple harmonic motion is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Substituting the values we have:
- \( m = 0.040 \, \text{kg} \)
- \( \omega = 10\pi \, \text{rad/s} \)
- \( A = 0.02 \, \text{m} \)

Calculating \( \omega^2 \) and \( A^2 \):
\[ \omega^2 = (10\pi)^2 = 100\pi^2 \]
\[ A^2 = (0.02)^2 = 0.0004 \, \text{m}^2 \]

Now substituting these into the energy formula:
\[ E = \frac{1}{2} \times 0.040 \times 100\pi^2 \times 0.0004 \]
\[ E = 0.0008 \times 100\pi^2 \]
\[ E = 0.08\pi^2 \]

### Step 5: Approximate the value of \( \pi^2 \)
Using \( \pi^2 \approx 10 \):
\[ E \approx 0.08 \times 10 = 0.8 \, \text{J} \]

### Step 6: Convert to millijoules
Since the answer options are in millijoules:
\[ E \approx 0.8 \, \text{J} = 800 \, \text{mJ} = 8 \times 10^{-3} \, \text{J} \]

### Final Answer
The total mechanical energy of the system is approximately:
\[ E \approx 7.9 \times 10^{-3} \, \text{J} \]