Question

If a seconds pendulum on the earth is taken to a planet whose gravity is half of the gravity on earth, its time period on that planet is

• A. 2 sec
• B. 4 sec
• C. \(4\sqrt{2}\) sec
• D. \(2\sqrt{2}\) sec

Hint:

Remember the definition of a "seconds pendulum" (its period is 2s, not 1s, because one "tick" or half-period is 1s). Also, recall the inverse square root relationship between period and gravity. Lower gravity means a longer (slower) period.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are considering a "seconds pendulum," which has a specific time period on Earth. We need to find its new time period on a planet with different gravity.

Step 2: Key Formula or Approach:
A

seconds pendulum is defined as a pendulum having a time period of exactly 2 seconds on Earth.
The formula for the time period (\(T\)) of a simple pendulum is:
\[ T = 2\pi\sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. From this formula, we can see that the time period is inversely proportional to the square root of gravity (\(T \propto \frac{1}{\sqrt{g}}\)).

Step 3: Detailed Explanation:
Let \(T_E\) and \(g_E\) be the time period and gravity on Earth.
Let \(T_P\) and \(g_P\) be the time period and gravity on the planet.
We are given:
- \(T_E = 2\) s (definition of a seconds pendulum).
- \(g_P = \frac{g_E}{2}\).
Using the proportionality \(T \propto \frac{1}{\sqrt{g}}\), we can set up a ratio:
\[ \frac{T_P}{T_E} = \frac{1/\sqrt{g_P}}{1/\sqrt{g_E}} = \sqrt{\frac{g_E}{g_P}} \]
Substitute the given relationship for gravity:
\[ \frac{T_P}{T_E} = \sqrt{\frac{g_E}{g_E/2}} = \sqrt{2} \]
Now, solve for the time period on the planet, \(T_P\):
\[ T_P = T_E \times \sqrt{2} \]
\[ T_P = 2 \times \sqrt{2} = 2\sqrt{2} \text{ sec} \]

Step 4: Final Answer:
The time period of the pendulum on the planet is \(2\sqrt{2}\) seconds.