The acceleration due to gravity on the surface of moon is 1.7 m s⁻². What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s⁻²)
The time period of the simple pendulum on the surface of moon is approximately 8.4 s.
Explanation
This question tests the relationship between time period of a simple pendulum and acceleration due to gravity. From the textbook formula T = 2π, we can see that time period is inversely proportional to the square root of acceleration due to gravity. Since the same pendulum is taken to the moon, its length L remains constant. By taking the ratio of time periods on earth and moon, the length cancels out, allowing us to find the unknown time period on the moon.
Solution Steps
Step 1: Write the formula for time period of a simple pendulum: T = 2π
Step 2: For the same pendulum, length L is constant. Therefore, T ∝ 1/√g, or T₁/T₂ =
Step 3: Substitute the given values: T_moon/T_earth =
Step 4: T_moon = T_earth × = 3.5 ×
Step 5: T_moon = 3.5 × .76 = 8.4 s