Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed oscillations [`nT_(1)=(n-1)T_(2)`, where `T_(1)` is time period of shorter length & `T_(2)` be time period of longer length and n are no. of oscillations completed]

Options

A. 5
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the number of oscillations \( n \) that the shorter pendulum completes when both pendulums are back in phase. We will use the time period formula for a simple pendulum and the given relationship between the oscillations of the two pendulums.

### Step-by-Step Solution

1. **Identify the lengths of the pendulums:**
- Length of the shorter pendulum, \( L_1 = 0.5 \, \text{m} \)
- Length of the longer pendulum, \( L_2 = 20 \, \text{m} \)

2. **Write the formula for the time period of a simple pendulum:**
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

3. **Calculate the time periods for both pendulums:**
- For the shorter pendulum:
\[ T_1 = 2\pi \sqrt{\frac{L_1}{g}} = 2\pi \sqrt{\frac{0.5}{g}} \]
- For the longer pendulum:
\[ T_2 = 2\pi \sqrt{\frac{L_2}{g}} = 2\pi \sqrt{\frac{20}{g}} \]

4. **Use the condition given in the problem:**
According to the problem, they will be in phase again when:
\[ nT_1 = (n-1)T_2 \]

5. **Substitute the time periods into the equation:**
\[ n \cdot 2\pi \sqrt{\frac{0.5}{g}} = (n-1) \cdot 2\pi \sqrt{\frac{20}{g}} \]

6. **Cancel \( 2\pi \) and \( \sqrt{g} \) from both sides:**
\[ n \sqrt{0.5} = (n-1) \sqrt{20} \]

7. **Square both sides to eliminate the square roots:**
\[ n^2 \cdot 0.5 = (n-1)^2 \cdot 20 \]

8. **Expand the equation:**
\[ 0.5n^2 = 20(n^2 - 2n + 1) \]
\[ 0.5n^2 = 20n^2 - 40n + 20 \]

9. **Rearrange the equation:**
\[ 0 = 20n^2 - 0.5n^2 - 40n + 20 \]
\[ 0 = 19.5n^2 - 40n + 20 \]

10. **Multiply through by 2 to eliminate the decimal:**
\[ 0 = 39n^2 - 80n + 40 \]

11. **Use the quadratic formula to solve for \( n \):**
The quadratic formula is given by:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 39 \), \( b = -80 \), and \( c = 40 \).
\[ n = \frac{80 \pm \sqrt{(-80)^2 - 4 \cdot 39 \cdot 40}}{2 \cdot 39} \]
\[ n = \frac{80 \pm \sqrt{6400 - 6240}}{78} \]
\[ n = \frac{80 \pm \sqrt{160}}{78} \]
\[ n = \frac{80 \pm 4\sqrt{10}}{78} \]

12. **Calculate the approximate value of \( n \):**
Since we are looking for a positive integer solution, we will consider the positive root:
\[ n \approx \frac{80 + 4\sqrt{10}}{78} \]

13. **Final calculation:**
Evaluating \( \sqrt{10} \approx 3.16 \):
\[ n \approx \frac{80 + 12.64}{78} \approx \frac{92.64}{78} \approx 1.19 \]
Thus, \( n \) can be approximated to 1.

### Conclusion
The number of oscillations \( n \) completed by the shorter pendulum when both pendulums are back in phase is approximately 1.