A particle moving along the x-axis is in a system that has pot...

Question

A particle moving along the x-axis is in a system that has potential energy U = x³ - 3x J, where x is in m. For each, is it a point of stable or unstable equilibrium?

Accepted Answer

Hello, let's go through this practice problem. Imagine a spaceship cruising along a spacetime continuum traveling on the X axis as it moves through space, it experiences a potential energy U equals X to the power of five minus X cubed joules where X represents the spaceship's position in meters determine if the spaceship is unstable or unstable equilibrium at each of its equilibrium points. Option A X equals 3/5 is the point of stable equilibrium and X equals negative 3/5 is the point of unstable equilibrium. Option B negative 3/5 and positive 3/5 for stable and unstable equilibrium respectively. Option C five thirds and negative 3/5 and option D negative 3/5 and negative five thirds. So this problem is asking us to find the equilibrium points for the spaceship. And so it's important to understand what that means. Recall that equilibrium points occur when the potential energy function is at its either minimum or its maximum. So the first thing we want to do is find where either minimums or maximums occur for the potential energy function. And to find this out, we're going to have to do some calculus because for maximizing or minimizing a function, we have to take the derivative of the function and then set that derivative equal to zero. So that is going to be our first step. We need to find the points where the derivative of the potential energy function with respect to X is equal to zero. So let's start with the potential energy function U equals X rays to the power of five minus X cubed and take the derivative of this function. So luckily for us, this is not a very difficult derivative to perform. We just need to apply the power rule for differentiation which states that we just need to take the exponent that is attached to each variable of differentiation multiply it by the entire term and then subtract that exponent by one. So let's begin with the X race of the power of five term. So that is an exponent of five. So we're going to multiply the entire term by five. But instead of raising the X to the power of five, we raise it to the power of four, which is one less than five. Now we're going to do the same thing for the X cubed term. So it's raise the power of three. So we multiply the entire term by three and then square the X rather than cubit. So now that we've found that derivative, now we need to set that derivative equal to zero and solve for X to find out which values are going to correspond. So five X raised the power of four is equal to three X squared, we can cancel out two of those XS. So we find that five X squared is equal to three. And then if we divide both sides of the equation by five and take the square root to solve four X, we find that X can be either positive or negative 3/5. So our equilibrium points are 3/5 and negative 3/5. However, the problem is asking us to figure out which one of them is the point of stable equilibrium and which one of them is the point of unstable equilibrium. So we do still have a bit more work to do the important rule. To remember here is that for stable equilibrium, the second derivative of the potential energy with respect to X will be greater than zero. Whereas for unstable equilibrium for unstable equilibrium, the second derivative of potential energy with respect to X will be less than zero, it will be negative. So let's go back to the function for potential energy that we discussed earlier. And now let's find its second derivative. So we'll just take the derivative and differentiate it again. So we're going to use the same power rule. We're gonna take the exponent and multiply it by the entire term. So four multiplied by five is 20 then we're going to reduce the exponent by one. So that's to the power of three then for the second term, we're gonna take the exponent of two and multiply it by three. So that is six multiplied by X, which is now S square, uh which is now has no square because two minus one is just one. So now, all we've got to do is plug in the two values for the equilibrium positions we found and see which one is positive and which one is negative. So let's begin with X equals 3/5 the positive one. So we're going to plug in to 20 multiplied by 3/5 raise the power of three minus six multiplied by 3/5. And if we put that into a calculator, then we find a value of ze 0.72 which is clearly greater than zero. So the th the positive 3/5 term must be the point of stable equilibrium. Now let's do the same thing for the negative 3/5 term. So that's two multiplied by negative 3/5 race, the power of three minus six multiplied by negative 3/5 put this into a calculator and we find negative 0.72 which is clearly less than zero. So X equals negative 3/5 must be the point of unstable equilibrium. So if we put this, so if positive 3/5 is stable and negative 3/5 is ne is unstable and that corresponds to option A. So A is our answer to the problem and that is it for this video. I hope this video helped you out and please consider checking out our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

A particle moving along the x-axis is in a system that has potential energy U = x³ - 3x J, where x is in m. For each, is it a point of stable or unstable equilibrium?

Key Concepts

Potential Energy

Equilibrium Points

Stability Analysis

Watch next

A particle moving along the x-axis is in a system that has potential energy U = x3 - 3x J, where x is in m. For each, is it a point of stable or unstable equilibrium?

Key Concepts

Potential Energy

Equilibrium Points

Stability Analysis

Watch next

A particle moving along the x-axis is in a system that has potential energy U = x³ - 3x J, where x is in m. For each, is it a point of stable or unstable equilibrium?