MCQ Detailed View

The correct answer is Increase. The time period of a simple pendulum depends on the length of the pendulum and the acceleration due to gravity (g) at the location where it swings. The formula for the time period $T$ of a simple pendulum is:

T=2πLgT = 2\pi \sqrt{\frac{L}{g}}T=2πgL

Where:

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  $T$ = time period (seconds)

  
  


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  $L$ = length of the pendulum (meters)

  
  


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  $g$ = acceleration due to gravity (m/s²)

  
  

On Earth, g≈9.8 m/s2g \approx 9.8 \, m/s^2g≈9.8m/s2. On the Moon, the gravitational acceleration is much smaller, approximately 1/6th of Earth’s gravity (gmoon≈1.63 m/s2g_{moon} \approx 1.63 \, m/s^2gmoon≈1.63m/s2).

Since the time period $T$ is inversely proportional to the square root of gravity, when the pendulum is taken to the Moon, the smaller value of $g$ causes $T$ to increase. This means the pendulum will swing slower and take more time to complete one oscillation.

For example, a pendulum that takes 2 seconds per swing on Earth will take approximately 4.8 seconds per swing on the Moon, assuming the same length. This behavior is due to the reduced gravitational pull on the Moon, which exerts less restoring force on the pendulum.

It is important to note that:

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  The length of the pendulum remains constant.

  
  


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  The pendulum cannot become zero or stop completely unless friction or external forces are removed.

  
  


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  The time period does not decrease because weaker gravity slows the motion rather than speeds it up.

  
  

🔑 Key Facts:

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  Time period T=2πL/gT = 2\pi \sqrt{L/g}T=2πL/g

  
  


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  Moon’s gravity gmoon≈1/6 gearthg_{moon} \approx 1/6 \, g_{earth}gmoon≈1/6gearth

  
  


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  Pendulum swings slower on the Moon.

  
  


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  Time period increases, not decreases.

  
  


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  Demonstrates the effect of gravity on oscillatory motion.

  
  

👉 Final Answer: When a pendulum is taken to the Moon, its time period increases.