The time period of a simple pendulum on a planet is 5 seconds and the effective length is 5 meters. How long should it be made to make 20 oscillation in 30 seconds ?

Question

Accepted Answer

To determine the correct length of the pendulum that allows it to make 20 oscillations in 30 seconds, we need to understand the relationship between the time period of a pendulum and its length. The time period $ T $ of a simple pendulum is given by the formula:

$$ T = 2\pi \sqrt{\frac{L}{g}} $$

where:
- $ T $ is the time period,
- $ L $ is the length of the pendulum,
- $ g $ is the acceleration due to gravity.

**Step-by-step Explanation:**

1. **Initial Conditions:**
   - Given that the time period $ T_1 $ on the planet is 5 seconds for a pendulum length $ L_1 = 5 $ meters.

2. **Calculate the Acceleration due to Gravity $ g $:**
   - Using the formula for the time period:
     $$ 5 = 2\pi \sqrt{\frac{5}{g}} $$
   - Solving for $ g $:
     $$ \frac{5}{2\pi} = \sqrt{\frac{5}{g}} $$
     $$ \left(\frac{5}{2\pi}\right)^2 = \frac{5}{g} $$
     $$ g = \frac{5 \times (2\pi)^2}{25} $$
     $$ g = \frac{4\pi^2}{5} $$

3. **New Conditions for 20 Oscillations in 30 Seconds:**
   - The time period $ T_2 $ for 20 oscillations in 30 seconds is:
     $$ T_2 = \frac{30}{20} = 1.5 \text{ seconds} $$

4. **Calculate the Required Length $ L_2 $:**
   - Using the formula for the time period with the new time period:
     $$ 1.5 = 2\pi \sqrt{\frac{L_2}{g}} $$
   - Solving for $ L_2 $:
     $$ \frac{1.5}{2\pi} = \sqrt{\frac{L_2}{g}} $$
     $$ \left(\frac{1.5}{2\pi}\right)^2 = \frac{L_2}{g} $$
     $$ L_2 = g \times \left(\frac{1.5}{2\pi}\right)^2 $$
     $$ L_2 = \frac{4\pi^2}{5} \times \frac{2.25}{4\pi^2} $$
     $$ L_2 = \frac{2.25}{5} $$
     $$ L_2 = 0.45 \text{ meters} $$

**Conclusion:**

- **Option 3 (0.45 m) is correct** because it matches the calculated length required for the pendulum to have a time period of 1.5 seconds, allowing it to complete 20 oscillations in 30 seconds.

**Explanation for Incorrect Options:**

- **Option 1 (1 m):** This length would result in a longer time period than 1.5 seconds, thus not allowing 20 oscillations in 30 seconds.
- **Option 2 (0.90 m):** This length is still too long and would result in a time period greater than 1.5 seconds.
- **Option 4 (2 m):** This length is significantly longer, resulting in an even longer time period, making it impossible to achieve 20 oscillations in 30 seconds.

Thus, Option 3 is the only correct choice based on the calculations and the conditions provided.