Professor Anderson

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>> Hello, class. Professor Anderson here. We've been talking about simple harmonic motion and what we did was we derived the solution to the differential equation that describes simple harmonic motion and remember the equation that we came up with was the following. x equals A cosine of omega t plus phi. This is the general solution for that differential equation. If I change phi to something else, I can turn a cosine into a sine, okay. Phi just really identifies the starting parameters of the problem. So, let's plot this out for the following. Let's just simplify, and we will let phi equal 0. And if phi equals 0, then x is A cosine omega t. What is V? V is the derivative of that, which we said is negative A omega sine of omega t. And acceleration is the derivative of that, which is negative A omega squared cosine omega t. Let's plot these out and see what they look like. This is x versus t. And it looks like a cosine. We know what a cosine looks like, it starts at some amplitude, a, goes down to negative a, goes back up, and so forth. Looks like that. That's a cosine. What about velocity and acceleration? Well, to do this carefully, let's identify these nodal points. And we will draw these other curves below. Okay. V is negative sine. We know that a positive sine would go up, like that. But a negative sine means that it has to go down. So it goes like this, and it reaches its maximum negative value when the position is going back to 0. This is where the block is going back through equilibrium. And now it's going to go up and reach its maximum value when the position is again 0, but the block is going the other way. And then it does it again. And so forth. What about the acceleration a? a is a cosine, but there's a negative sign in front. All right. that's pretty easy to draw, because we have a cosine right here. So all I have to do is flip it over. And so if I flip that over, it's going to look like this. Okay. This is what the position, the velocity and the acceleration look like as a function of time. And you know what these values are here, it goes up to A, it goes down to negative A. This one goes from negative A omega up to positive A omega. And this one goes from negative A omega squared to positive A omega squared. It's just what's ever in the front of those things. And so let's just be explicit. X max is, of course, A. V max, in terms of amplitude, don't worry about the sine, it's omega A. And a max is omega squared times A. Let's look at the case where phi is no longer 0 and see if we can determine phi from the starting parameters. Hello, class. Let's go back to simple harmonic motion for just one second. Let's go back to our general solution, X equals A cosine omega t plus phi. And let's say that you know something about the starting parameters. Let's say that you are given the starting position, X i. And you're also given the starting speed, or velocity, V i. How can we determine phi from those? Well, the way we do it is the following. Starting conditions means that t equals 0. And that means that X initial, according to this equation, is A cosine of omega times 0 plus phi. So this just becomes a cosine phi. Initial velocity we know was negative A omega sine of omega t plus phi. But we're going to let that, again, equal 0, and so we get negative A omega sine of phi. And so I have two equations here. I have V i, which equals negative A omega sine phi, and I have X i, which equals A cosine phi. And anytime I have two equations, I can always divide one equation by the other. So let's take this equal sign out of here. Let's divide one equation by the other, and look what happens. The A drops out and we get V i over X i equals negative sine phi over cosine phi, but sine over cosine is tangent of phi. And so there's this very nice way to determine phi. You write tangent of phi is negative V i over, and we forgot something here, right? We had an omega sitting right there. Let's get that back in. Negative omega over omega times X i. So, if you're given the starting condition, V i and X i, you know omega, all you have to do is take the arc tangent of that stuff and you can determine phi.

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>> Hello, class. Professor Anderson here. We've been talking about simple harmonic motion and what we did was we derived the solution to the differential equation that describes simple harmonic motion and remember the equation that we came up with was the following. x equals A cosine of omega t plus phi. This is the general solution for that differential equation. If I change phi to something else, I can turn a cosine into a sine, okay. Phi just really identifies the starting parameters of the problem. So, let's plot this out for the following. Let's just simplify, and we will let phi equal 0. And if phi equals 0, then x is A cosine omega t. What is V? V is the derivative of that, which we said is negative A omega sine of omega t. And acceleration is the derivative of that, which is negative A omega squared cosine omega t. Let's plot these out and see what they look like. This is x versus t. And it looks like a cosine. We know what a cosine looks like, it starts at some amplitude, a, goes down to negative a, goes back up, and so forth. Looks like that. That's a cosine. What about velocity and acceleration? Well, to do this carefully, let's identify these nodal points. And we will draw these other curves below. Okay. V is negative sine. We know that a positive sine would go up, like that. But a negative sine means that it has to go down. So it goes like this, and it reaches its maximum negative value when the position is going back to 0. This is where the block is going back through equilibrium. And now it's going to go up and reach its maximum value when the position is again 0, but the block is going the other way. And then it does it again. And so forth. What about the acceleration a? a is a cosine, but there's a negative sign in front. All right. that's pretty easy to draw, because we have a cosine right here. So all I have to do is flip it over. And so if I flip that over, it's going to look like this. Okay. This is what the position, the velocity and the acceleration look like as a function of time. And you know what these values are here, it goes up to A, it goes down to negative A. This one goes from negative A omega up to positive A omega. And this one goes from negative A omega squared to positive A omega squared. It's just what's ever in the front of those things. And so let's just be explicit. X max is, of course, A. V max, in terms of amplitude, don't worry about the sine, it's omega A. And a max is omega squared times A. Let's look at the case where phi is no longer 0 and see if we can determine phi from the starting parameters. Hello, class. Let's go back to simple harmonic motion for just one second. Let's go back to our general solution, X equals A cosine omega t plus phi. And let's say that you know something about the starting parameters. Let's say that you are given the starting position, X i. And you're also given the starting speed, or velocity, V i. How can we determine phi from those? Well, the way we do it is the following. Starting conditions means that t equals 0. And that means that X initial, according to this equation, is A cosine of omega times 0 plus phi. So this just becomes a cosine phi. Initial velocity we know was negative A omega sine of omega t plus phi. But we're going to let that, again, equal 0, and so we get negative A omega sine of phi. And so I have two equations here. I have V i, which equals negative A omega sine phi, and I have X i, which equals A cosine phi. And anytime I have two equations, I can always divide one equation by the other. So let's take this equal sign out of here. Let's divide one equation by the other, and look what happens. The A drops out and we get V i over X i equals negative sine phi over cosine phi, but sine over cosine is tangent of phi. And so there's this very nice way to determine phi. You write tangent of phi is negative V i over, and we forgot something here, right? We had an omega sitting right there. Let's get that back in. Negative omega over omega times X i. So, if you're given the starting condition, V i and X i, you know omega, all you have to do is take the arc tangent of that stuff and you can determine phi.