Two identical positive charges Q each are placed on the x axis at point (-a,0) and (a,0) A point charge of magnitude q is placed at the origin. For small displacement along x axis, the charge q executes simple harmonic motion if it is poitive and its time period is `T_(1)`. if the charge q is negative, it perform oscillations when displaced along y axis. in this case the time period of small oscillations is `T_(2).` find `(T_(1))/(T_(2))`

Answer

To solve the problem, we need to analyze the forces acting on the charge \( q \) when it is displaced along the x-axis and y-axis. We will derive the time periods \( T_1 \) and \( T_2 \) for the respective cases and then find the ratio \( \frac{T_1}{T_2} \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical positive charges \( Q \) located at points \( (-a, 0) \) and \( (a, 0) \).
- A point charge \( q \) is placed at the origin \( (0, 0) \).
- We will analyze the forces acting on \( q \) when it is displaced along the x-axis and y-axis.

2. **Case 1: Positive Charge \( q \) Displaced Along the x-axis**:
- When the charge \( q \) is displaced by a small distance \( x \) along the x-axis, the distances from the charges become \( A + x \) and \( A - x \).
- The forces due to the two charges can be expressed as:
\[ F_1 = \frac{kQq}{(a+x)^2}, \quad F_2 = \frac{kQq}{(a-x)^2} \]
- The net force \( F \) acting on \( q \) can be calculated as:
\[ F = F_1 - F_2 = \frac{kQq}{(a+x)^2} - \frac{kQq}{(a-x)^2} \]
- Using the binomial approximation for small \( x \) (i.e., \( (a+x)^{-2} \approx \frac{1}{a^2} - \frac{2x}{a^3} \)), we can simplify the expression for \( F \):
\[ F \approx -\frac{4kQq}{a^3} x \]
- This indicates that the motion is simple harmonic with:
\[ F = -k_{\text{eff}} x \quad \text{where } k_{\text{eff}} = \frac{4kQq}{a^3} \]
- The angular frequency \( \omega \) is given by:
\[ \omega^2 = \frac{k_{\text{eff}}}{m} = \frac{4kQq}{ma^3} \]
- Therefore, the time period \( T_1 \) is:
\[ T_1 = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{ma^3}{4kQq}} = \frac{2\pi}{\sqrt{4kQq}} \sqrt{ma^3} \]

3. **Case 2: Negative Charge \( q \) Displaced Along the y-axis**:
- When the charge \( q \) is displaced by a small distance \( y \) along the y-axis, the distance from the charges becomes \( \sqrt{a^2 + y^2} \).
- The forces acting on \( q \) due to the two charges can be expressed as:
\[ F = 2 \cdot \frac{kQ(-q)}{(a^2 + y^2)} \cdot \frac{y}{\sqrt{a^2 + y^2}} \quad \text{(only vertical components add)} \]
- For small \( y \), we can approximate \( \sqrt{a^2 + y^2} \approx a \):
\[ F \approx -\frac{2kQq}{a^3} y \]
- This also indicates simple harmonic motion with:
\[ k_{\text{eff}} = \frac{2kQq}{a^3} \]
- The angular frequency \( \omega \) is given by:
\[ \omega^2 = \frac{2kQq}{m a^3} \]
- Therefore, the time period \( T_2 \) is:
\[ T_2 = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi \sqrt{\frac{ma^3}{2kQq}} = \frac{2\pi}{\sqrt{2kQq}} \sqrt{ma^3} \]

4. **Finding the Ratio \( \frac{T_1}{T_2} \)**:
- Now we can find the ratio:
\[ \frac{T_1}{T_2} = \frac{\frac{2\pi}{\sqrt{4kQq}} \sqrt{ma^3}}{\frac{2\pi}{\sqrt{2kQq}} \sqrt{ma^3}} = \frac{\sqrt{2}}{\sqrt{4}} = \frac{1}{\sqrt{2}} \]

### Final Answer:
\[ \frac{T_1}{T_2} = \frac{1}{\sqrt{2}} \]