A simple pendulum 2.00 m long swings through a maximum angle of 30.0° with the vertical. Calculate its period (a) assuming a small amplitude, and (b) using the first three terms of Eq. (14.35). (c) Which of the answers in parts (a) and (b) is more accurate? What is the percentage error of the less accurate answer compared with the more accurate one?

Video transcript

Hey, everyone in this problem, we're told that the equation of simple harmonic motion of a simple pendulum is derived using assumptions that make it valid for small displacements. During an experiment, we suspend a spherical ball using a 1.2 m long thread and displace the pendulum by 28 degrees were asked to calculate the period of the pendulum assuming small displacement. And using this other equation were given where t the period equal to two pi times the square root of L over G times one plus one squared over two squared sine squared data over two plus one squared times three squared over two squared times four squared times sine to the exponent four of data over to, okay, we're asked to state which value is more accurate and to determine the percentage error in the other value by comparing it with the value we considered accurate. Alright, so that's a lot to digest. Um let's start with this period, were asked to find the period of the pendulum using small displacement and recall that the period of a pendulum for small displacements is given by T is equal to two pi times the square root of L over G okay. Alright. So part one is asking us which value is more accurate? Okay. So we have this equation that we've just written out where we have, we assume that we have small displacement. And then we have this longer equation given in the problem that has those sine squared signed to the four terms. Okay. Which one is more accurate? Well, the longer equation is going to be more accurate. Okay, it has more sign terms, okay. So it has more accuracy. There's more information in this equation compared to this shorter equation, two pi times the square root of L over G we've truncated that you can see we have times essentially one and we've ignored these extra sign terms in that expansion. Okay. So we have more terms, more information, more accuracy in this longer equation. And when you're given equations, and one of them is telling you that you have to make an assumption. So for example, in this case, to use the equation T is equal to two pi times the square root of L over G, we have to assume that we have small displacement that typically means the equation is less accurate. Okay. If you have to make an assumption like that, typically the equation is an approximation. Alright. So for part one which one is more accurate or saying that the period calculated from the given equation, the equation that has the sine squared and signed to the fourth term. Alright. And if we look at our answer choices that already eliminates choice B choice C and choice F, Which leaves us with three choices. So now let's calculate the period with each of these equations okay. In order to determine that percentage error, so starting with the approximate period okay, where we assume a small displacement, We have two pi times the square root of L over G. That's gonna be two pi times the square root in this problem, the length of the thread L is given by 1.2 m. G is a gravitational acceleration which we know is 9. meters per second squared. Okay. We have meters divided by meters per second squared inside of this square root. That means that inside the square root we're gonna have a unit of seconds squared. And then when we take the square root that's gonna be seconds, which is a unit for period we want. So that works out when we get a value of 2. seconds. And now using that other equation that we were given the period T is going to be equal to. And let me just put a box around this other value we found first so that we can keep track of it. Okay. So we have T is equal to two pi times the square root of L over G. Okay. It starts the same way And then we multiply by one plus 1/2 squared, which is going to be four Sine squared of theta over two plus, we have one squared times three squared. So that's gonna be nine divided by two squared, which is four times four squared, which is 16, so nine divided by four times 16, all times signed to the exponent four of theta over two. That's the equation we're given. Now, in this problem, data is 28°. We're told that the pendulum is displaced by 28°. And so theta over to the value that we use inside of that function is going to be 14°. And so when we calculate T we have T is equal to two pi times the square root of L 1. m divided by G the gravitational acceleration 9.8 m per second squared, just like in the period calculation for the small displacement. And then we multiply by one plus 1/ Sine squared of 14° plus 9/ signed to the exponent for of 14 degrees. And if we work all of this out, we are going to get a value. Let me give us some more room to work here. We are going to get a value of 2.23189 seconds approximately. Okay. And that is our period using that more accurate equation. So we have two values for the period. And now we're asked to find the percentage error. Okay. So for part two, we're asked to find the percentage error and recall that for the percentage error, the equation is going to be her son error is equal to the absolute value of the true value or the more accurate value minus the approximate value all divided by the true value. And because we want a percentage, we multiply this by 100. So in our case, the true value is the one that is the more accurate equation, which is the second equation that has a sine squared and signed to the four terms. So that's going to be 2.23189 seconds or approximate value is the one that we had to assume small displacement. And what we found there was we had a period of 2. seconds. Okay. If we go back up, that was the other value that we squared in blue 2.198657 seconds. And we're gonna divide that by the true value 2.23189 seconds. And then multiply it all by 100. This is going to give us a value of 0.01 or 89, approximately times 100 for a percent error of approximately 1.489%. Okay. And so that is part two, that percent error of 1.489%. If we go back to our answer choices, we've already eliminated three based on part one. Now we're, we see that the answer choices are rounded to the nearest percent. We have 1.489%. So to the nearest percent that's going around to 1%. And so we get answer choice. A okay t calculated from the given equation is more accurate and our percent error using the other equation is approximately 1%. Thanks everyone for watching. I hope this video helped see you in the next one.