A body executes simple harmonic motion under the action of a force `F_1` with a time period `(4)/(5)s` . If the force is changed to `F_(2)`, it executes SHM with time period `(3)/(5)s`. If both the forces `F_(1) and F_(2)` act simultaneously in the same direction on the body, its time period (in seconds) is

Options

A. `(12)/(25)`
B. `(24)/(25)`
C. `(35)/(24)`
D. `(25)/(12)`

Answer

To solve the problem, we need to find the time period of a body executing simple harmonic motion (SHM) when two forces, \( F_1 \) and \( F_2 \), act simultaneously. Given the time periods associated with each force, we can use the relationship between the time period and the force.

### Step-by-Step Solution:

1. **Understand the relationship between time period and force**:
The time period \( T \) of a body in SHM is inversely proportional to the square root of the force acting on it. This can be expressed mathematically as:
\[ T \propto \frac{1}{\sqrt{F}} \]
or
\[ T^2 \propto \frac{1}{F} \]

2. **Set up the equations for the two forces**:
Let \( T_1 \) be the time period for force \( F_1 \) and \( T_2 \) be the time period for force \( F_2 \).
- Given \( T_1 = \frac{4}{5} \) seconds for force \( F_1 \)
- Given \( T_2 = \frac{3}{5} \) seconds for force \( F_2 \)

3. **Write the equations for the forces**:
From the relationship, we can express the forces in terms of the time periods:
\[ F_1 = k \cdot \frac{1}{T_1^2} \quad \text{and} \quad F_2 = k \cdot \frac{1}{T_2^2} \]
where \( k \) is a constant.

4. **Calculate the forces**:
- For \( F_1 \):
\[ F_1 = k \cdot \frac{1}{\left(\frac{4}{5}\right)^2} = k \cdot \frac{25}{16} \]
- For \( F_2 \):
\[ F_2 = k \cdot \frac{1}{\left(\frac{3}{5}\right)^2} = k \cdot \frac{25}{9} \]

5. **Find the combined force**:
When both forces act simultaneously, the total force \( F \) is:
\[ F = F_1 + F_2 = k \cdot \frac{25}{16} + k \cdot \frac{25}{9} \]
To combine these, we need a common denominator, which is 144:
\[ F = k \left(\frac{25 \cdot 9}{144} + \frac{25 \cdot 16}{144}\right) = k \cdot \frac{25(9 + 16)}{144} = k \cdot \frac{25 \cdot 25}{144} = k \cdot \frac{625}{144} \]

6. **Calculate the new time period \( T \)**:
Using the relationship \( T^2 \propto \frac{1}{F} \):
\[ T^2 = \frac{k}{F} = \frac{k \cdot 144}{625} \]
Since \( T_1^2 = \frac{k}{F_1} \) and \( T_2^2 = \frac{k}{F_2} \), we can find the new time period \( T \):
\[ T = \sqrt{\frac{144}{625}} \cdot T_1 = \sqrt{\frac{144}{625}} \cdot \frac{4}{5} \]
\[ T = \frac{12}{25} \cdot \frac{4}{5} = \frac{48}{125} \text{ seconds} \]

### Final Answer:
The time period of the body when both forces \( F_1 \) and \( F_2 \) act simultaneously is \( \frac{48}{125} \) seconds.