A “seconds” pendulum has a period of exactly 2.000 s. That is,...

Question

A “seconds” pendulum has a period of exactly 2.000 s. That is, each one-way swing takes 1.000 s. What is the length of a seconds pendulum in Austin, Texas, where g = 9.793 m /s² ? If the pendulum is moved to Paris, where g = 9.809 m/s², by how many millimeters must we lengthen the pendulum? What is the length of a seconds pendulum on the Moon, where g = 1.62 m/s² ?

Accepted Answer

Welcome back. Everyone in this problem. A physics professor is on a tour. He is to deliver lectures in various places for the lectures. He is carrying along a pendulum with him. He wants the pendulum to have a period of exactly one second to four significant figures. In Florida. The professor measures the value of G to be 9.798 m per second squared. And in Anchorage, it turns out to be 9.819 m per second squared. On the other hand, the value of G on MARS is 3.721 m per second squared, determine the length of the professor's pendulum in Florida. And the length adjustment in millimeters he makes going from, from Florida to Anchorage to the pendulum's length. Also find the length of the pendulum on Mars. If the period is to be the same for our answer choices, it tells us the length in Florida is 2.48 multiplied by 10 to the negative 1 m that they, it increases by 5.319 multiplied by 10 to the negative 4 m. And the length in Mar is 9.425 multiplied by 10 to the negative 2 m. B tells us the length in Florida is 2.007. The length in adjustment increases by 2.148 multiplied by 10 to the negative 3 m. And the length in Mars is 9.425 multiplied by 10 to the negative 2 m. C as the length in Florida is 2.482 multiplied by 10 to the negative 1 m. The length in adjustment decreases by 5.319 multiplied by 10 to the negative 4 m. And the length in Mars is 9.425 multiplied by 10 to the negative 2 m. And the D tells us the length in far is 2.007 m. The length in adjustment decreases by 2.148 multiplied by 10 to the negative 3 m. And the length in Mars is 9.425 multiplied by 10 to the negative 2 m. Now, in this problem, we want to find the length of the pendulum and we want to find the length of the pendulum, both in Florida and Mars and in Anchorage. And we need it for Anchorage to figure out the length adjustment he'd make going from Florida to Anchorage. So if we're going to figure all of that out, we need a way to find the length of the pendulum at any point. How can we relate the length of a pendulum to its period. Well, recall that the period T is equal to two pi multiplied by the square root of the length divided by our acceleration due to gravity G. No, if we were to solve for L here, the lens that we want to find, then we can square both sides to get T squared equal to four pi squared, Ls sorry, four pi squared L not L squared because it was under the square root four pi squared L OK, divided by G. And now to solve four L, we can multiply both sides by G and divide both sides by four pi squared to get L to be equal to GT squared divided by four pi squared. Now this is the length at any point. So now let's use that to find the length given our acceleration due to gravity in Florida. OK? In Anchorage. And on Mars, let's start with Florida and let's call that LF subscript F to denote that we're talking about Florida. So for Florida, OK, then the length of the pendulum is going to be equal to the acceleration due to gravity there 9.798 m per second squared multiplied by the period squared. Remember that it was one second, OK? Divided by four multiplied by pi squared. And that lens OK is going to be equal to 0.24819 m. Next, let's find the length for Anchorage So for Anchorage again, we use the same formula, let's denote it as L A but no, OK. It's going to be equal to the acceleration due to gravity there, which is 9.819 m per second squared, multiplied by our period squared. OK. Divided by four pi squared. And that length is going to be equal to 0.24872 m. Finally, let's do it for Mars. Now for Mars, let's call that LM and its and its acceleration due to gravity, there is 3.721 meters per second squared. So we're gonna multiply that by one squared and then divide by four pi squared. And again, when we do that, then we get the length on Mars to be equal to 0.094254 m. So the length of the pendulum, as we can see depends on our acceleration due to gravity and the period. No, we've figured out a part of our question. We've figured out the length in Florida. Oops, not, not that one, the length in Florida. OK. We've also figured out the length on Mars. The last part of our problem tells us to figure out the length adjustment that the prof professor makes going from Florida to anchor it. OK. So now to figure that out, we can subtract the length from Anchorage. Uh sorry, we can subtract the length at Florida from the length at Anchorage and put that into our formula to solve for its length. OK. So now if we put that here, then the length adjustment, the change in length I should write, write that nearby. So the length adjustment OK is going to be a change in lens which is L Anchorage minus the length in Florida. And now to figure that out, we can just put that difference in length at our point for G because it is our acceleration due to gravity that changes to figure it out. So now our, our, our value for G is going to be 9.819 m per second squared minus 9.798 m per second squared. All of that is multiplied by the period one second. And then it's all divided by four by squared. And now again, when you solve for that, then you should get the change in length to be equal to 5.31934 multiplied by 10 to the negative 4 m. No note that the length in Anchorage is greater than the length it was in Florida. So that means the length should be increased. OK. So to four, no, now that we have all of our answers approximating them to four significant figures. OK? Then the length in Florida is going to be approximately 2.482 multiplied by 10 to the negative 1 m. The length sorry, the change in length or the length adjustment is going to be equal to 5.319 multiplied by 10 to the negative 4 m. And the length on Mars is going to be approximately equal to. I should write these as approximately approximately equal to 9.425 multiplied by 10 to the negative 2 m. If we look back at our answer choices that tells us, then that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Pendulum Motion

Acceleration due to Gravity (g)

Length Adjustment for Pendulum

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