Question:medium

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

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For a pendulum, \(T\propto \frac{1}{\sqrt g}\). If \(g\) becomes half, time period becomes \(\sqrt2\) times.
  • \(2\) sec
  • \(4\) sec
  • \(4\sqrt2\) sec
  • \(2\sqrt2\) sec
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
A "seconds pendulum" is a pendulum whose time period is exactly 2 seconds on Earth. The time period of a simple pendulum depends on its length and the local acceleration due to gravity. When the pendulum is moved to another planet, its length remains the same, but gravity changes, which in turn changes its time period.
Step 2: Key Formula or Approach:
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 the formula, we can see the relationship $T \propto \frac{1}{\sqrt{g}}$. We can use this proportionality to find the new time period. \[ \frac{T_{planet}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{planet}}} \] Step 3: Detailed Explanation:
Given: - Time period on Earth, $T_{earth} = 2$ s (definition of a seconds pendulum). - Gravity on the planet, $g_{planet} = \frac{1}{2} g_{earth}$. Using the ratio formula: \[ \frac{T_{planet}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{planet}}} = \sqrt{\frac{g_{earth}}{\frac{1}{2}g_{earth}}} = \sqrt{2} \] Now, solve for the time period on the planet: \[ T_{planet} = T_{earth} \times \sqrt{2} \] \[ T_{planet} = 2 \times \sqrt{2} = 2\sqrt{2} \text{ sec} \] Step 4: Final Answer:
The time period of the pendulum on the new planet is $2\sqrt{2}$ seconds. Therefore, option (D) is correct.
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