![]() ![]() And now can- if you tell me atĥ.8 seconds, where is x, I can tell you.\): A block is attached to a spring and placed on a frictionless table. Sophisticated calculus, solved a differential equation. The motion of the block on a spring in SHM is defined by the position x (t) Acos t + ) with a velocity of v (t) A sin ( t + ). Right, because the amplitude of this cosine function is A-Ī cosine- and instead of writing w, we can now write the While staying constant, the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring: ETotal U + K 1 2kx2 + 1 2mv2. We were right about the A, and that's just intuition, So now we've figured out theĪctual function that describes the position of that springĪs a function of time. We said that this differentialĮquation is true, if this is x of t, and omega isĮqual to this. It, if omega squared is equal to k over m. Squared- or omega squared, I think that's omega. True if what is true? This equation holds true if mw ![]() Have mw squared cosine of omega t is equal to kĬosine of omega t. Minus mAw squared cosine of wt is equal to minus Second derivative here, and I substituted the function here. And then I took the secondĭerivative of it. ![]() Is x of t, just based on our intuition of the drawing. Which is essentially aĭifferential equation, I just rewrote acceleration as Said that by the spring constant, if you rewrite forceĪs mass times acceleration, you get this. I have some space, without getting rid of that nice curve I I don't want to get rid of this,īecause I think this gives us some intuition Substituted x of t, which I guess that's that, in here. Prime, the second derivative, into this, and I just Times my original function- times x of t. T, which is in this case is this, times minus Aw So if this is true, I shouldīe able to say that m times the second derivative of x of But the minus is still there,īecause I had the minus to begin with. Of- these are just scalar values, right? These are just constants. Let me do this in a differentĬolor, just so it doesn't get monotonous. And then, if we want the secondĭerivative- so that's x prime prime of t. Just doing the chain rule- the derivative of cosine of t is Inside, so it'll be that omega, times the To A cosine of wt, what is the derivative of this? Let's test this expression- this function- to see if it To go into that just yet, we'll get a little bit more T, where this is the angular velocity of- well, I don't want Us that it's a cosine function, with amplitude A. Intuition for the position over time- our intuition tells What the position is, at any given time, of this massĪttached to the spring. We're going to use that to guessĪt what a solution to this differential equation is. We're going to use our intuitionīehind what we did earlier in the previous video. Get your head around it, but this is as good an example asĮver to be exposed to it. But the solution to differentialĮquations is actually going to be a function, We've done in the past are numbers, essentially, or a set And the solution to aĭifferential equation isn't just a number, right? A solution to equations that This, you not only have a function, but you haveĭerivatives of that function. In one expression, or in one equation, on both sides of I've just written, this is actually a differential Notation, just so you remember this is a function of time. That's just the secondĭerivative of x as a function of t. I think the easiest notation would just be x prime prime. We know- that acceleration is the secondĭerivative of x as a function of t. Keep all of this, just so we remember what we're talkingĪbout this whole time. Twice of x of t, right? So let's rewrite this equation And acceleration is theĭerivative of velocity, or the second derivative of position. Saying, well if x is a function of t, what'sĭerivative of x with respect to time, right? Your change in position Video, I'd just rewritten the spring equation. Once you know the answer by experiment, see if you can figure out why the answer is what it is. So if the period is the same on jupiter, that means it will also be the same in an accelerating elevator, and if it is different on jupiter, that means it will also be different in the elevator. You can tell this by figuring out what your weight would be in the elevator - if you work it out you will see it will be m(g+a). Now how does this apply to acceleration? When you are in an elevator accelerating up at a rate of a, that is exactly the same as if gravity increased from g to g+a. Go ahead and do that and see if gravity makes a difference to the period. Now look over on the right side and you will see that you can change gravity from that of Earth to that of Jupiter. To measure the period, measure the time for 10 or 20 bounces and divide by 10 or 20. You can measure the period of oscillation by clicking on the box for "stopwatch". Here is a simulation of a mass on a vertical spring ![]()
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