ASVAB (Armed Services Vocational Aptitude Battery)Mechanical Comprehension (MC)Hard
A simple pendulum of length 1 meter has a period of oscillation of approximately 2 seconds on Earth. If this pendulum were taken to a planet where the acceleration due to gravity is four times that of Earth, what would be its new period of oscillation?
- A1 second
- B4 seconds
- C2 seconds
- D0.5 seconds
Show answer & explanationAnswer & explanation
Correct answer: A. 1 second
The period (T) of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length and g is the acceleration due to gravity. If g is quadrupled (g' = 4g), the new period T' would be T' = 2π√(L/4g) = 2π * (1/√4) * √(L/g) = (1/2) * [2π√(L/g)]. So, T' = T/2. If the original period was 2 seconds, the new period would be 2/2 = 1 second.
Why the other options are wrong
- B. Incorrect. This would imply gravity was reduced by a factor of 4, or period was directly proportional to g.
- C. Incorrect. This would imply gravity had no effect on the period, or that it was the same as Earth.
- D. Incorrect. This would be if gravity was 16 times Earth's gravity, or if period was proportional to 1/g.
Simple Pendulum Period
The time taken for a simple pendulum to complete one full oscillation, dependent on its length and the acceleration due to gravity.
- T = 2π√(L/g)
- Period is independent of mass and amplitude (for small angles).
- Increasing length increases period; increasing gravity decreases period.
Memory trick: Lengthy Loops, Gravity Greatly Governs!