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. We know the time periods for each force acting alone, \( T_1 \) and \( T_2 \).

### Step-by-Step Solution:

1. **Identify the Given Time Periods**:
- The time period for force \( F_1 \) is \( T_1 = \frac{4}{5} \) seconds.
- The time period for force \( F_2 \) is \( T_2 = \frac{3}{5} \) seconds.

2. **Relate Time Period to Angular Frequency**:
- The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
- Therefore, we can express \( \omega_1 \) and \( \omega_2 \) for the two forces:
\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{\frac{4}{5}} = \frac{10\pi}{4} = \frac{5\pi}{2} \]
\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{\frac{3}{5}} = \frac{10\pi}{3} \]

3. **Calculate the Effective Angular Frequency**:
- When both forces act simultaneously, the effective angular frequency \( \omega \) can be found using the formula:
\[ \frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} \]
- First, calculate \( T_1^2 \) and \( T_2^2 \):
\[ T_1^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
\[ T_2^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \]

4. **Substitute into the Equation**:
- Now substitute these values into the equation:
\[ \frac{1}{T^2} = \frac{1}{\frac{16}{25}} + \frac{1}{\frac{9}{25}} = \frac{25}{16} + \frac{25}{9} \]

5. **Find a Common Denominator**:
- The common denominator of 16 and 9 is 144:
\[ \frac{25}{16} = \frac{225}{144}, \quad \frac{25}{9} = \frac{400}{144} \]
- Therefore,
\[ \frac{1}{T^2} = \frac{225 + 400}{144} = \frac{625}{144} \]

6. **Calculate \( T^2 \)**:
- Taking the reciprocal gives:
\[ T^2 = \frac{144}{625} \]

7. **Find \( T \)**:
- Finally, take the square root to find \( T \):
\[ T = \sqrt{\frac{144}{625}} = \frac{12}{25} \text{ seconds} \]

### Final Answer:
The time period when both forces \( F_1 \) and \( F_2 \) act simultaneously is \( \frac{12}{25} \) seconds.

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