A body of mass 20 g connected to a spring of spring constant k, executes simple harmonic motion with a frequency of `(5//pi)` Hz. The value of spring constant is

Options

A. `4Nm^(-1)`
B. `3Nm^(-1)`
C. `2Nm^(-1)`
D. `5Nm^(-1)`

Explanation

To find the spring constant \( k \) for a body executing simple harmonic motion, we can use the formula relating frequency \( f \), mass \( m \), and spring constant \( k \).

### Step-by-step Solution:

1. **Identify the given values:**
- Mass \( m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \)
- Frequency \( f = \frac{5}{\pi} \, \text{Hz} \)

2. **Use the formula for frequency in terms of mass and spring constant:**
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]

3. **Rearranging the formula to solve for \( k \):**
\[ k = (2\pi f)^2 m \]

4. **Substituting the values into the equation:**
- First, calculate \( 2\pi f \):
\[ 2\pi f = 2\pi \left(\frac{5}{\pi}\right) = 10 \]
- Now substitute \( 10 \) into the equation for \( k \):
\[ k = (10)^2 \cdot 0.02 \]

5. **Calculate \( k \):**
\[ k = 100 \cdot 0.02 = 2 \, \text{N/m} \]

### Conclusion:
The value of the spring constant \( k \) is \( 2 \, \text{N/m} \).

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