A simple pendulum oscillates with an amplitude of $10.0 ^\circ$. What fraction of the time does it spend between $+5.0 ^\circ$ and $-5.0 ^\circ$? Assume SHM.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. I wrote this equation for a pendulum just because it's really close looking, there are similar looking to this equation for a spring. So, we have the restoring force is approximately equal to negative mg θ for a pendulum. And for a spring the restoring force is, you know, negative kx, Hooke's law. And when you see the similarities between these you can see that θ is like x, and mg is like k but that doesn't really matter. The important thing is that θ is like x. And so any equation that we have for x which is position as a function of time such as this equation, we can write the letter x there or we can write the letter θ there instead. This is true if you can assume the pendulum motion to be simple harmonic motion, SHM which the question tells us to do. So, θ is like x and that means θ as a function of time then is the amplitude which is θ max times cos of 2π over the period times time. OK. So, θ max times cos 2π over period times time equals some position. And this position is going to be 5 degrees or negative 5 degrees. And, but anyway, we'll get there in a second. So, this is divided by θ max here. And you have cos 2π over period times t is some whatever θ you like divided by θ max and the thetas that we like are positive or negative 5 degrees and so, initially, this pendulum starts at θ max because the cosine function starts at its maximum, goes through 0 and then back up to 0 and back up to θ max. So, here's θ max, and I'm graphing the cosine function as a function of time. And here's 5 degrees, and here's negative 5 degrees. And the question is, for what period of time what fraction of time is the pendulum between 5 and negative 5 degrees during a full cycle? So, that's going to be this period of time plus. And then, so it goes beyond negative 5 and then back to negative 5, plus this period of time. So, those 2 squiggled portions of time divided by this full period is going to be the answer to our question. It's pretty abstract. So, I don't know how much sense that's making. But anyway, hopefully that's in some sense. So, if I were to, you know, of course that you don't see any pendulums here. So, maybe I should draw a pendulum to show how that's related to this graph. We have a pendulum here. And it starts at θ max that corresponds to this time here, time 0. And then it's swinging, and then it hits 5 degrees and then crosses over to negative 5 degrees. And this portion of the swaying corresponds to this bit of time here. And the time is short because it's going at its maximum velocity here, right? Maximum velocity is right at the bottom. So, it passes from 5 to negative 5 quite quickly. And then it goes to its its furthest extent on the other side of the swing. So, it goes all the way to θ max again here, 10 degrees. And then slowly starts to swing back and then it hits this negative 5 degrees again. And then we count this portion of time as it goes from negative 5 degrees back to positive 5 degrees. Alrighty. So, our job is to figure out at what times does the pendulum cross these thresholds of 5 degrees and negative 5 degrees. So, we'll solve this equation for positive 5 degrees which is positive 1/2 because 5 over 10 is a half. And then we'll solve it again for negative 5 degrees. So, having a negative 1/2 here. So, take the inverse cosine of both sides and 2π over period times t1. This is gonna be the first time it passes 5 degrees. It's gonna be the inverse cosine of 1/2, and we have a special triangle which says the invers cosine of 1/2 is π over 3. And it's also good to think about trig functions in terms of a unit circle, especially since we have an obtuse angle as our answer for the next portion here, for t2 and on the unit circle you have a point being, having X coordinate cosine and Y coordinate sine of θ. And... So, when you have... And the reason that drawing this unit circle is to show you where the next solution is going to be. Because when we take, I mean, let's say that this part here, let's suppose that you're familiar with this special triangle and you're happy that inverse cosine of 1/2 is π over 3, in which case, you know, 2π over period times t1 equals π over 3, in which case t1 is T over 6. And so that's fine. And then the next question is, when does... At what time t2 do you have inverse cosine of negative 1/2. Well, negative 1/2 occurs at this point, where the cosine is I guess more like this point here, where the cosine is negative 1/2 and that happens at this angle here and that's going to be the reference angle which is π over 3 taken away from this full straight line angle of π. Yeah, there's a lot going on here this is like a more like a a question for your math course than it is for physics. But, you know, physics involves math. So, there you go. So, the inverse cosine of negative 1/2. What I'm trying to say is that it's π minus this reference angle in here which is 2/3 π. And I think actually if you plug that in your calculator it'll tell you that anyway as a decimal. Let's find out. I'm curious how smart is your calculator. I think it will do that. So, inverse cosine of negative 1/2 is... That didn't work because I wanted to be in radians. Inverse cosine of negative 1/2 is that. No, this looks good, 2/3 π. OK. Excellent. So yeah, you can just ignore this whole unit circle business because this question is simple enough that your calculator is just going to give you the correct answer. So, yeah t2 is a 2 T over 6 which is T over 3. But you could have some decimal number there and that would be fine too. So, the fraction of time spent between 5 and negative 5 degrees is going to be the time taken to go from t2 to t1 which is this first portion. Here's t1 and here's t2. So, it's going to be that difference in time there. Multiplied by 2 because that time interval occurs again on the back swing of the pendulum. So, t1 minus t2 multiplied by 2 divided by the period is a fraction of time spent between 5 and negative 5 degrees. So is 2 times T over 3 minus T over 6 and the T's cancel everywhere. It's 2 times 2 6 minus the 6 which is 2 times 2 over 6. 1/3. 1/3 is the answer here folks. That's the amount of time the pendulum spends between negative 5 and 5 degrees.