A particle is executing simple harmonic motion under the action of a force `F` with a time period `(3)/(5)`s. When the force is changed to `F'`, the time period of oscillation is `(4)/(5)`s. When both the forces `F` and `F'` act simultaneously in the same direction on the body, time period in seconds in `T = (6a)/(5b)`. COmpute the value of a + b.

Explanation

To solve the problem, we need to determine the time period of oscillation when two forces \( F \) and \( F' \) act simultaneously on a particle executing simple harmonic motion (SHM). We are given the following information:

1. The time period \( T_1 \) under force \( F \) is \( \frac{3}{5} \) seconds.
2. The time period \( T_2 \) under force \( F' \) is \( \frac{4}{5} \) seconds.
3. When both forces act together, the time period is given as \( T_3 = \frac{6a}{5b} \).

### Step-by-step Solution:

1. **Identify the relationship between time period and angular frequency**:
The time period \( T \) of a particle in SHM is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Therefore, we can express the angular frequencies corresponding to \( T_1 \) and \( T_2 \):
\[ \omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{\frac{3}{5}} = \frac{10\pi}{3} \]
\[ \omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{\frac{4}{5}} = \frac{10\pi}{4} = \frac{5\pi}{2} \]

2. **Use the relationship of angular frequencies when forces are combined**:
When both forces \( F \) and \( F' \) act simultaneously, the relationship between their angular frequencies is given by:
\[ \omega_3^2 = \omega_1^2 + \omega_2^2 \]

3. **Calculate \( \omega_1^2 \) and \( \omega_2^2 \)**:
\[ \omega_1^2 = \left(\frac{10\pi}{3}\right)^2 = \frac{100\pi^2}{9} \]
\[ \omega_2^2 = \left(\frac{5\pi}{2}\right)^2 = \frac{25\pi^2}{4} \]

4. **Combine the squares of the angular frequencies**:
To combine \( \omega_1^2 \) and \( \omega_2^2 \), we need a common denominator:
\[ \omega_1^2 + \omega_2^2 = \frac{100\pi^2}{9} + \frac{25\pi^2}{4} \]
The common denominator of 9 and 4 is 36. Thus, we convert both fractions:
\[ \frac{100\pi^2}{9} = \frac{400\pi^2}{36} \]
\[ \frac{25\pi^2}{4} = \frac{225\pi^2}{36} \]
Now, add them:
\[ \omega_3^2 = \frac{400\pi^2}{36} + \frac{225\pi^2}{36} = \frac{625\pi^2}{36} \]

5. **Find \( T_3 \)**:
Now, we find \( T_3 \):
\[ \omega_3 = \sqrt{\frac{625\pi^2}{36}} = \frac{25\pi}{6} \]
Therefore, the time period \( T_3 \) is:
\[ T_3 = \frac{2\pi}{\omega_3} = \frac{2\pi}{\frac{25\pi}{6}} = \frac{12}{25} \]

6. **Equate \( T_3 \) to the given expression**:
We know from the problem that:
\[ T_3 = \frac{6a}{5b} \]
Setting this equal to our calculated \( T_3 \):
\[ \frac{12}{25} = \frac{6a}{5b} \]

7. **Cross-multiply to find \( a \) and \( b \)**:
Cross-multiplying gives:
\[ 12 \cdot 5b = 25 \cdot 6a \]
Simplifying:
\[ 60b = 150a \quad \Rightarrow \quad 2b = 5a \quad \Rightarrow \quad b = \frac{5}{2}a \]

8. **Choose values for \( a \) and \( b \)**:
If we let \( a = 2 \), then:
\[ b = \frac{5}{2} \cdot 2 = 5 \]

9. **Calculate \( a + b \)**:
Finally, we compute \( a + b \):
\[ a + b = 2 + 5 = 7 \]

### Final Answer:
The value of \( a + b \) is \( 7 \).