Two identical springs are fixed at one end and masses `1kg` and `4kg` are suspended at their other ends. They are both strethced down form their mean position and let go simultaneously. If they are in the same phase after every `4` seconds then the springs constant `k` is

Options

A. `pi(N)/(m)`
B. `pi^(2)(N)/(m)`
C. `2pi(N)/(m)`
D. given data is insufficient

Explanation

To solve the problem, we need to find the spring constant \( k \) given that two masses (1 kg and 4 kg) suspended from identical springs are in the same phase after every 4 seconds.

### Step-by-Step Solution:

1. **Identify the Masses and Their Corresponding Angular Frequencies:**
- Let \( m_1 = 1 \, \text{kg} \) and \( m_2 = 4 \, \text{kg} \).
- The angular frequency \( \omega \) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
- Therefore, for mass \( m_1 \):
\[ \omega_1 = \sqrt{\frac{k}{1}} = \sqrt{k} \]
- For mass \( m_2 \):
\[ \omega_2 = \sqrt{\frac{k}{4}} = \frac{1}{2}\sqrt{k} \]

2. **Determine the Condition for Same Phase:**
- The two systems will be in the same phase when the difference in their angular frequencies multiplied by time is an integer multiple of \( 2\pi \):
\[ \omega_1 t - \omega_2 t = 2n\pi \]
- Substituting the expressions for \( \omega_1 \) and \( \omega_2 \):
\[ \sqrt{k} t - \frac{1}{2}\sqrt{k} t = 2n\pi \]
- Simplifying this gives:
\[ \left(\sqrt{k} - \frac{1}{2}\sqrt{k}\right)t = 2n\pi \]
\[ \frac{1}{2}\sqrt{k} t = 2n\pi \]

3. **Using Given Time Interval:**
- We know that they are in the same phase every 4 seconds, so let \( t = 4 \) seconds:
\[ \frac{1}{2}\sqrt{k} \cdot 4 = 2n\pi \]
- This simplifies to:
\[ 2\sqrt{k} = 2n\pi \]
- Dividing both sides by 2:
\[ \sqrt{k} = n\pi \]

4. **Finding the Spring Constant \( k \):**
- For \( n = 1 \) (the first occurrence of being in phase):
\[ \sqrt{k} = \pi \]
- Squaring both sides gives:
\[ k = \pi^2 \]

### Final Answer:
The spring constant \( k \) is:
\[ \boxed{\pi^2} \, \text{N/m} \]