A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where g = 3.71 m/s^2?

Video transcript

Hey everyone welcome back in this problem. A plumb bob displaced by a small angle from the equilibrium position has a period of 1.23 seconds on the Earth. Okay. And we're asked to find the period of this simple pendulum when taken to the surface of planet K. For the gravitational acceleration G is 13. m/s. Alright, so we have some information about the period on earth. We want to find information about the period on a certain planet and were given information about the gravitational acceleration. So let's recall. We can relate the period of a pendulum to the gravitational acceleration through the following two pi over the period. T. Is equal to the square root of G over L. The gravitational acceleration divided by that length of the pencil. Alright, so we want to find the period T on planet K. And we know the gravitational acceleration G. On that planet, we're told it but we don't know the length. L. What we do know is information on Earth on Earth, we know the period and we know the gravitational acceleration that's going to allow us to find the length L. Okay? And then we can use that to help us find the period on planet K. Okay? Because the length of the pendulum doesn't change on one planet to the other. So let's start on Earth And we have two pi divided by the period on Earth, we're told is 1.23 seconds this is going to equal the square root of the gravitational acceleration. And again we're on Earth, this is 9.8 m per second squared divided by the length L. Which we're looking for. Alright, so we want to solve for L. We can square both sides. Two pi divided by 1.23 seconds is equal to 9.8 m per second squared, divided by the length out. And if we rearrange L is going to be equal to 9. m per second squared Times 1.23 seconds. And sorry, we missed the square here in the last step. Okay, this should be two pi divided by 1.23 seconds. All squared. Okay, because we're squaring to remove this square root. Okay, that's better. So when we rearrange we get 1.23 seconds divided by two pi All squared. There we go. And if we work this out on your calculator you're going to get the length is gonna be 0. m. Alright, so we found the length of the pendulum on earth, and again the length of the pendulum on earth and the length of the pendulum on planet K are going to be the same. So now we can use this in our equation for planet K to find the period. Alright, so going to Planet K and again that same equation, two pi divided by the period T. And in this case this is the period T we're looking for is equal to the square root of the gravitational acceleration, which we're told is 13.2 m per second squared on planet K. Divided by L. The length of the pendulum, which we just found to be 0. m. Alright. So if we isolate for T we get that T. Is equal to two pi times the square root. 0. m divided by 13.2 m per second squared. Ok. The unit meter will cancel. We're going to end up with the unit of second squared in the square root and then when we square root we will have the unit of second that we want for period. Okay. Alright. And if we work this out, we're gonna get a period T approximately 1. seconds on planet K. So if we go back up to our answer choices, we see that the period of this pendulum on planet K is going to be D 1.6 seconds. Thanks everyone for watching. I hope this video helped see you in the next one