A particle of mass 200 g executes a simpel harmonit motion. The resrtoring force is provided by a spring of spring constant `80 N m^-1`. Find the time period.

Options

A. 0.93 s
B. 0.63 s
C. 0.31 s
D. None of these

Explanation

To find the time period of a particle executing simple harmonic motion (SHM) attached to a spring, we can use the formula:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( T \) is the time period,
- \( m \) is the mass of the particle in kilograms,
- \( k \) is the spring constant in Newtons per meter (N/m).

### Step-by-Step Solution:

1. **Convert Mass to Kilograms**:
The mass given is 200 grams. To convert grams to kilograms, we use the conversion factor \( 1 \text{ kg} = 1000 \text{ g} \):
\[ m = \frac{200 \text{ g}}{1000} = 0.2 \text{ kg} \]

2. **Identify the Spring Constant**:
The spring constant \( k \) is given as \( 80 \text{ N/m} \).

3. **Substitute Values into the Formula**:
Now we substitute the values of \( m \) and \( k \) into the formula for the time period:
\[ T = 2\pi \sqrt{\frac{0.2 \text{ kg}}{80 \text{ N/m}}} \]

4. **Calculate the Fraction Inside the Square Root**:
First, calculate \( \frac{0.2}{80} \):
\[ \frac{0.2}{80} = \frac{1}{400} = 0.0025 \]

5. **Calculate the Square Root**:
Now we find the square root:
\[ \sqrt{0.0025} = 0.05 \]

6. **Multiply by \( 2\pi \)**:
Now we multiply by \( 2\pi \):
\[ T = 2\pi \times 0.05 = 0.1\pi \]

7. **Approximate \( \pi \)**:
Using \( \pi \approx 3.14 \):
\[ T \approx 0.1 \times 3.14 = 0.314 \text{ seconds} \]

8. **Final Result**:
Therefore, the time period \( T \) is approximately \( 0.314 \) seconds. Rounding to two decimal places, we get:
\[ T \approx 0.31 \text{ seconds} \]

### Conclusion:
The time period of the particle executing simple harmonic motion is **0.31 seconds**.