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Periodicity of algebraic models

Sal analyzes the periodicity of graphs that model real world situations.

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  • blobby green style avatar for user InnocentRealist
    It has occurred to me that the sine could be defined in terms of periodic motion of a spring, where F = -kx is proportional to the negative of the displacement, and results in a periodic displacement function. I suspect that it has been proven at some point in the past that this is the same function as the sine (the y coordinate) in the unit circle, since this is how we calculate the displacement of an oscillating spring + weight. However, how do we know they are the same? What is the proof of this?
    Intuitively it seems that there is a simple relationship between these two kinds of motion, circular and oscillating.

    With gravity, acceleration is constant, velocity is a linear function v = at + s, and distance is quadratic.
    With an oscillating spring, acceleration is a linear function of and in the opposite direction of displacement. But the displacement itself oscillates, so the acceleration also oscillates. Then so does the velocity. I imagine all 3 of these are sine functions.
    Is there an exponential function (with a common factor of displacement, or something) somehow involved in this?
    (4 votes)
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  • blobby green style avatar for user John Murphy
    I used to think of 2Pi as a constant approximately 6.28 but I'm beginning to realize is that 2Pi of a Sin or Cos function can be greater or less that 6.28 of whatever unit be used. For example if we squeeze a sinusoidal function down we can to less than 2Pi as a hard constant of 6.28. Well then it really is 2Pi because it represents a full revolution or a full cycle, but not necessarily 6.28 on the x axis. So how should Pi and 2Pi be regarded? Is it just a measurement of rotation?
    (1 vote)
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  • hopper cool style avatar for user Ralph Turchiano
    Just out of curiosity, if this was a very rough COS function would it look like -16.35 (PI/30X)+17.85 ? Going from a min height of (0, 1.7) to a max height of (30, 34). plus or minus
    (0 votes)
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  • blobby green style avatar for user Binda Satpathy
    This is a sine curve and we know that 1 revolution is at 2 pi, so shouldn't 1 revolution of the Ferris wheel be at 90 secs? At 60 secs, he would have only made half the revolution. Please clarify.
    (0 votes)
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    • stelly blue style avatar for user Kim Seidel
      I don't know where you got 90 secs for 1 revolution. You have to look at the graph. At time 0, the ferris wheel is at the bottom of it's rotation. At time = 30 seconds, the ferris wheel is at its max height (this would be 1/2 revolution). At time = 60, the ferris wheel is back at the bottom if its rotation (so 1 full rotation has now been completed). The rotation of the ferris wheel will be a function of it's size and its speed.
      (7 votes)

Video transcript

- [Voiceover] We're told Divya is seated on a Ferris Wheel at time t equals zero. The graph below shows her height h in meters, t seconds after the ride starts. So at time equals zero, she is, looks like about two, what is this, this would be one and a half, so it looks like she's about two meters off the ground and then as time increases, she gets as high as, it looks like this is close to 30, maybe 34 meters and then she comes back down, looks like two meters and up to 34 meters again, so let's read the question. So the question asks us, approximately how long does it take Divya to complete one revolution on the Ferris Wheel? All right, so this is interesting, so this is when she's at the bottom of the Ferris Wheel, so then she gets to the top of the Ferris Wheel, and then she keeps rotating until she gets back to the bottom of the Ferris Wheel again. So it took her 60 and t is in terms of seconds. So it took her 60 seconds to go from the bottom to the bottom again, and in another 60 seconds, she would have completed another revolution. And so let me fill that in, it is going to take her 60 seconds, 60 seconds, and we of course can check our answer if we like. Let's do another one of these. So here we have, a doctor observes the electrical activity of Finn's heart over a period of time. The electrical activity of Finn's heart is cyclical, as we hope it would be, and peaks every 0.9 seconds. Which of the following graphs could model the situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? Over here it looks like we peaked at zero seconds, and then here we're peaking a little bit more than one, this looks maybe at 1.1, maybe at 2.2 and 3.3. This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out A. This one is peaking, it looks like the interval between peaks is less than a second, but it looks like a good bit less than a second. It looks like maybe every three quarters of a second, or maybe every four-fifths of a second. Not quite nine-tenths, nine-tenths this first peak would be a little bit closer to one, but this one is close. Choice C is looking good. The first we're at zero, then the first peak, this looks pretty close to one but it's less than one. It looks like a tenth less than one, so I like choice C. Now choice D, it looks like we're peaking every half-second, so it's definitely not that. So this looks like a peak of every 0.9 seconds. This is the best representation that I... This is the best representation that I can think of. And you can actually verify that. If you have a peak every 0.9 seconds, you're going to have four peaks in 3.6 seconds. So one, two, three, four, this looks like it's at 3.6. Over here, you have one, two, three, four, you've had four peaks in less than three seconds. So this one definitely isn't 0.9. So instead of just even forcing yourself to eyeball just between this peak and that peak, you can say well, if we're every 0.9 seconds, how long would three peaks take or four peaks, and then you can actually get a little bit more precise as you try to eyeball it. So we can check our answer and verify that we got it right.