A system has potential energy U(x)=(10 J)[1−sin((3.14 rad/m) x)] as a particle moves over the range 0 m≤x≤3 m
b. For each, is it a point of stable or unstable equilibrium?

Question

A system has potential energy U(x)=(10 J)[1−sin((3.14 rad/m) x)] as a particle moves over the range 0 m≤x≤3 m 
b. For each, is it a point of stable or unstable equilibrium?

Accepted Answer

Welcome back, everyone. We are taking a look at a system in which a particle of mass is moving along. And we are told that its position represented by X sticks between zero and two m. We are given its potential energy function by this equation right here. And we are tasked with finding uh first, what are the equilibrium points within the system? And then second, determining the stability of each of our points. Let's go through our answer choices here. Before diving into the question, we have a, a stable equilibrium point at X equals zero and an unstable equilibrium point at X equals 1.41 B unstable equilibrium point at X equals zero and stable equilibrium point at X equals 1.41 C stable equilibrium point at X equals one and an unstable equilibrium point at X equals two or D unstable equilibrium point at X equals two and a stable equilibrium point at X equals one. Well, first and foremost, in order to find our equilibrium points, what we are going to do is we are going to take the derivative of our potential energy function with respect to X. And then we will set it equal to zero and this will give us our equilibrium points. So here's how we're going to calculate this. We have the derivative of, I'm gonna distribute this eight right here to the inside of the parentheses. So we have eight minus eight Os sign times 2.22 X wonderful. And so what this equals is eight, well, first derivative of this first eight and the eighth just a constant. So that disappears. Now, the derivative of negative eight cosine or two of 2.22 X. So this is going to be eight sign of 2.22 X because that's the derivative of the outside uh times the derivative of the inside, which is just 2.22 giving us 17.76 times the sign of 2.22 X wonderful. Now, remember we are going to set this equal to zero to find our equilibrium points. Now, first and foremost, what I can do is I can divide both sides by 17.76 and it just gets rid of it and zero divided by anything is zero. Now, what I'm gonna do is I'm going to take the a sign of both sides. Now, what does this give us? Well, the arc sign of zero is going to be N pi where N is any integer? If you think about the unit circle here, right, any integer times pi And if you take the sign of that will give you zero. So then we have N P is equal to 2.22, X. Finally, we just divide both sides by 2.222 point 22. And we get that X is equal to N pi over 2.22. And remember this is for all integers, we have zero, comma one, comma two, comma three and so on. So here's what we're going to do to find our equilibrium points. Remember our limitation here is that the position of the actual particle is going to be between zero and two inclusive. So we'll just plug in one end value at a time and see if it falls within that range. So plugging in zero here for N, we get zero times pi over 2.22 and this gives us zero for one of our equilibrium points. And let's go ahead and plug in one for N. We have that, this is just pi divided by 2.22 which gives us 1.41, go ahead and take one more step further. And we'll plug in two for N, this gives us two pi over 2.22, which is equal to 2.83. And as you'll notice 2.83 is outside of our range, so we can't use those two values. So let me just scroll down here just a little bit. We have that our equilibrium points then are zero comma 1.41 wonderful. Now, moving on to part two here, I'm gonna scroll down just a little bit more, moving on to part two. Here we are tasked with determining the stability of each of our equilibrium points. And as a reminder, the derivative with respect to X of our potential energy function was 17.76 times the sign of 2.22 X. Now how do we determine the stability here? Well, what we are going to do is we are going to take the second derivative of our potential energy function with respect to X. And then we are going to see when we plug in our equilibrium points, whether we get positive or negative or positive is going to correspond to a stable equilibrium point and negative is going to correspond to an unstable equilibrium point. So before plugging in our equilibrium points, let's first find this second derivative and we can do that by just taking the derivative of our first derivative. So the derivative of our first derivative gives us our desired second derivative of our potential energy function. And this is going to be 17.76 times the cosine of 2.22 X. That's the derivative of the outside according to the chain rule, take the derivative of the inside once again. So this will just be 2.22 which gives us 39.48 times the cosine of 2.22 X. Wonderful, great. So let me go ahead and scroll down here just a little bit. So now what we are going to do is we are going to take our equilibrium points and we are going to plug it into our second derivative equation. And remember positive corresponds to stable and negative corresponds to unstable. So first and foremost, let's see what happens when X is equal to zero on X equals zero. We have 39.48 times the cosine of 2.22 times zero. This is just equal to 39.48 times the cosine of zero and the cosine of zero is going to be one. So this gives us 39.48, a 39.48 is greater than zero, which means that it's positive and therefore the equilibrium point of zero, let me go ahead and erase my previous markings here. Our equilibrium point of zero is going to be positive and therefore stable. In fact, instead of positive, I will write stable up here as well. Wonderful. Now, let's see what happens when X is equal to 1.41, we have 39.48 times the cosine of 2.22 times 1.41, you plug this into your calculator, you actually get negative 39.44. Now this is less than zero. So this means that our equilibrium point of 1.41 is unstable, unstable. So now that we have found our equilibrium points and determined their stability. Let's go see which one of our answer choices corresponds to that answer here. All right. So answer choice A says that we have a stable equilibrium point at X knott and an unstable equilibrium point at 1.41, which is exactly what we found. So our final answer choice is answer choice. A thank you all so much for watching. I hope this video helped. We will see you all in the next one.

A system has potential energy U(x)=(10 J)[1−sin((3.14 rad/m) x)] as a particle moves over the range 0 m≤x≤3 m b. For each, is it a point of stable or unstable equilibrium?

Verified Solution

Key Concepts

Potential Energy

Equilibrium Points

Stability Analysis