You pull a simple pendulum 0.240 m long to the side through an angle of 3.50° and release it. (b) How much time does it take if the pendulum is released at an angle of 1.75° instead of 3.50°?

Video transcript

Hey everyone in this problem. We construct a 0.84 m long simple pendulum Using a light string and displace it by 4.3° on one side. Before releasing it to oscillate. Next you double the displacement angle to 8.6° on that same side. The more asked what is the difference between the two time periods for the two angles used? Alright, so we're asked about the period T. Were given some information about the length. L. So let's recall that we can relate those to the following two pi over the period. T is equal to the square root of G over L. Alright. So in order to calculate the period T we need to use G. The gravitational acceleration and L the length. Okay, so we could do that. The gravitational acceleration is going to be 9.8 m per second squared the length. L. Is 0.84 m long. And what you'll notice is that this does not depend on the angle, this does not depend on the displacement angle. So whether we displace it 4.3 degrees or 8.6 degrees, the period T is going to be the same. It depends only on the gravitational acceleration and the length of our pendulum. So the period does not change between the first displacement in the second displacement and so the difference between those two periods for those two different angles is going to be c zero seconds. That's it for this one. Thanks everyone for watching. See you in the next video