A simple pendulum with bob of mass m and length x is held in position at an angle `theta_(1)` and then angle `theta_(2)` with the vertical. When released from these positions, speeds with which it passes the lowest postions are `v_(1)&v_(2)` respectively. Then ,`(v_(1))/(v_(2))` is.

Options

A. `(1-costheta_(1))/(1-costheta_(2))`
B. `sqrt((1-costheta_(1))/(1-costheta_(2)))`
C. `sqrt((2gx(1-costheta_(1)))/(1-costheta_(2)))`
D. `sqrt((1-costheta_(1))/(2gx(1-costheta_(2))))`

Answer

To solve the problem, we will use the principle of conservation of energy. The potential energy lost by the pendulum bob when it is released will be converted into kinetic energy at the lowest point of its swing.

### Step-by-Step Solution:

1. **Identify the Initial and Final States**:
- The pendulum bob is initially at an angle θ₁ and θ₂ with the vertical.
- When released, it swings down to the lowest point where its speed is v₁ and v₂ respectively.

2. **Calculate the Height (h)**:
- The height (h) from which the bob falls can be calculated using the length of the pendulum (L) and the angle (θ).
- The vertical height fallen when the pendulum is at angle θ is given by:
\[ h = L - L \cos \theta = L(1 - \cos \theta) \]

3. **Apply Conservation of Energy**:
- The loss in gravitational potential energy (PE) when the bob falls to the lowest point is equal to the gain in kinetic energy (KE).
- The potential energy lost is:
\[ \Delta PE = mg \cdot h = mg \cdot L(1 - \cos \theta) \]
- The kinetic energy gained at the lowest point is:
\[ KE = \frac{1}{2} mv^2 \]
- Setting these equal gives:
\[ mgL(1 - \cos \theta) = \frac{1}{2} mv^2 \]

4. **Solve for Speed (v)**:
- Canceling mass (m) from both sides and rearranging gives:
\[ v^2 = 2gL(1 - \cos \theta) \]
- Taking the square root:
\[ v = \sqrt{2gL(1 - \cos \theta)} \]

5. **Calculate Speeds v₁ and v₂**:
- For angle θ₁:
\[ v_1 = \sqrt{2gL(1 - \cos \theta_1)} \]
- For angle θ₂:
\[ v_2 = \sqrt{2gL(1 - \cos \theta_2)} \]

6. **Find the Ratio of Speeds**:
- To find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \frac{\sqrt{2gL(1 - \cos \theta_1)}}{\sqrt{2gL(1 - \cos \theta_2)}} \]
- Simplifying gives:
\[ \frac{v_1}{v_2} = \sqrt{\frac{1 - \cos \theta_1}{1 - \cos \theta_2}} \]

### Final Answer:
\[ \frac{v_1}{v_2} = \sqrt{\frac{1 - \cos \theta_1}{1 - \cos \theta_2}} \]