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The given problem describes a coupled system of two pendulums where the ratio of time periods for in-phase and out-of-phase oscillations is 2:1. We need to calculate the force constant \( k \) of the coupling spring.

### Step-by-Step Solution:

1. **In-Phase and Out-of-Phase Oscillations**: 
   The time periods for in-phase (\( T_{\text{in}} \)) and out-of-phase (\( T_{\text{out}} \)) oscillations are related by:
   \[
   \frac{T_{\text{in}}}{T_{\text{out}}} = 2
   \]
   Given the ratio is 2:1, we have \( T_{\text{in}} = 2T_{\text{out}} \).

2. **Time Period of Simple Pendulum**:
   The time period of a single simple pendulum is given by:
   \[
   T_0 = 2\pi \sqrt{\frac{l}{g}}
   \]
   where:
   - \( l = 1.5 \, \text{m} \) is the length of the rod,
   - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

3. **Time Periods for the Coupled System**:
   - For **in-phase oscillation**, the time period is approximately the same as that of an individual pendulum:
     \[
     T_{\text{in}} \approx T_0
     \]
   - For **out-of-phase oscillation**, the coupling spring modifies the time period. The time period for out-of-phase motion is given by:
     \[
     T_{\text{out}} = 2\pi \sqrt{\frac{m}{k}}
     \]
     where \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) is the mass of each pendulum, and \( k \) is the spring constant.

4. **Relation Between Time Periods**:
   From the given ratio:
   \[
   \frac{T_{\text{in}}}{T_{\text{out}}} = \frac{T_0}{T_{\text{out}}} = 2
   \]
   Thus,
   \[
   T_{\text{out}} = \frac{T_0}{2}
   \]

5. **Substituting into the Formula**:
   Using the formula for \( T_{\text{out}} \):
   \[
   T_{\text{out}} = 2\pi \sqrt{\frac{m}{k}}
   \]
   we can equate it to \( \frac{T_0}{2} \), giving:
   \[
   \frac{T_0}{2} = 2\pi \sqrt{\frac{m}{k}}
   \]
   Squaring both sides:
   \[
   \left(\frac{T_0}{2}\right)^2 = \left(2\pi \sqrt{\frac{m}{k}}\right)^2
   \]
   Simplifying:
   \[
   \frac{T_0^2}{4} = 4\pi^2 \frac{m}{k}
   \]
   Solving for \( k \):
   \[
   k = \frac{16\pi^2 m}{T_0^2}
   \]

6. **Substituting \( T_0 \)**:
   We already know:
   \[
   T_0 = 2\pi \sqrt{\frac{l}{g}}
   \]
   Squaring \( T_0 \):
   \[
   T_0^2 = 4\pi^2 \frac{l}{g}
   \]
   Substituting into the expression for \( k \):
   \[
   k = \frac{16\pi^2 m}{4\pi^2 \frac{l}{g}} = 4mg \frac{1}{l}
   \]
   Finally, substituting the known values \( m = 0.05 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( l = 1.5 \, \text{m} \):
   \[
   k = \frac{4 \times 0.05 \times 9.8}{1.5}
   \]
   Calculating:
   \[
   k = 1.307 \, \text{N/m}
   \]

### Final Answer:
The force constant of the coupling spring is approximately **1.31 N/m**.

Would you like further details or have any questions about the solution?

Here are some related questions to explore:
1. What is the difference between in-phase and out-of-phase oscillations?
2. How does coupling between pendulums affect their oscillation periods?
3. What factors influence the time period of a pendulum?
4. How would changing the mass of the pendulums affect the force constant \( k \)?
5. Can this method be applied to other types of coupled oscillators, such as masses on springs?

**Tip:** Always double-check the unit conversions in such problems to avoid small errors in calculations!

The ratio of time period of oscillations in in-phase oscillation and out-of-phase mode oscillation of a coupled system of two pendulums (via a spring) is 2:1. If the mass of each pendulum is 50 gm and each is suspended by massless rods of length 1.5 m each, find the force constant of the coupling spring.

In this problem, we determine the force constant of the coupling spring in a coupled system of two pendulums. The time period ratio for in-phase and out-of-phase oscillations is given as 2:1, with pendulums of mass 50g each, suspended by massless rods of length 1.5m. Learn how to calculate the force constant using oscillation formulas and harmonic motion principles. Suitable for Grades 11-12.

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