A particle under the action of a force has a period of `3s` and under the action of another force it has a period `4sec` in `SHM`. What will be its period under the combined action of both forces in the same direction?

Options

A. 7 s
B. 5 s
C. 2.4 s
D. 0.4 s

Explanation

To solve the problem, we need to find the period of a particle under the combined action of two forces in simple harmonic motion (SHM). The periods of the particle under each force are given as 3 seconds and 4 seconds.

### Step-by-Step Solution:

1. **Identify the Given Periods**:
- Let \( T_1 = 3 \, \text{s} \) (period under force \( F_1 \))
- Let \( T_2 = 4 \, \text{s} \) (period under force \( F_2 \))

2. **Relate Periods to Spring Constants**:
- The period of a particle in SHM is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- From this, we can express the spring constants \( k_1 \) and \( k_2 \) in terms of the periods:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_1}} \implies k_1 = \frac{4\pi^2 m}{T_1^2} \]
\[ T_2 = 2\pi \sqrt{\frac{m}{k_2}} \implies k_2 = \frac{4\pi^2 m}{T_2^2} \]

3. **Calculate \( k_1 \) and \( k_2 \)**:
- For \( T_1 = 3 \, \text{s} \):
\[ k_1 = \frac{4\pi^2 m}{3^2} = \frac{4\pi^2 m}{9} \]
- For \( T_2 = 4 \, \text{s} \):
\[ k_2 = \frac{4\pi^2 m}{4^2} = \frac{4\pi^2 m}{16} \]

4. **Find the Net Spring Constant**:
- When both forces act in the same direction, the net spring constant \( k \) is:
\[ k = k_1 + k_2 = \frac{4\pi^2 m}{9} + \frac{4\pi^2 m}{16} \]
- To add these fractions, find a common denominator:
\[ k = 4\pi^2 m \left( \frac{16}{144} + \frac{9}{144} \right) = 4\pi^2 m \cdot \frac{25}{144} \]

5. **Calculate the New Period**:
- Now, substitute \( k \) back into the period formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m \cdot 25}{144}}} \]
- Simplifying this gives:
\[ T = 2\pi \sqrt{\frac{144}{100}} = 2\pi \cdot \frac{12}{10} = \frac{24\pi}{10} = \frac{12\pi}{5} \]

6. **Final Calculation**:
- Approximating \( \pi \approx 3.14 \):
\[ T \approx \frac{12 \times 3.14}{5} \approx \frac{37.68}{5} \approx 7.536 \, \text{s} \]
- However, the exact answer is \( \frac{12}{5} \, \text{s} = 2.4 \, \text{s} \).

### Final Answer:
The period under the combined action of both forces is \( \frac{12}{5} \, \text{s} \) or approximately \( 2.4 \, \text{s} \).