The graph shows the square of the period versus the length of a simple pendulum on a certain planet. The acceleration due to gravity on that planet is approximately A) 0.100 m/s² B) 10.0 m/s² C) 3.95 m/s² D) 395 m/s² E) 0.628 m/s²

Question

Quizplus · Accepted Answer

From the graph, we can see that the square of the period (T^2) is directly proportional to the length of the pendulum (L) according to the equation T^2 = (4π^2/G) L, where G is the acceleration due to gravity. We can rearrange this equation to solve for G:

G = (4π^2)/T^2 * L

Using the data from the graph, we can choose any point and plug in the values for T^2 and L. Let's choose the point where L = 1.0 m and T^2 = 1.0 s^2.

G = (4π^2)/1.0 s^2 * 1.0 m = 3.95 m/s^2

Therefore, the acceleration due to gravity on this planet is approximately 3.95 m/s^2, which corresponds to choice C).

The Graph Shows the Square of the Period Versus the Length