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Sinusoidal Motion

  1. May 31, 2010 #1
    Can someone please explain the following equation:

    x = A sin(wt)

    where A = amplitude, w = angular velocity, t = time.

    The way it's explained in my textbook is very confusing.
  2. jcsd
  3. May 31, 2010 #2


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    Well what in the equation is confusing you exactly?

    Sinusoidal motion is basically a periodic motion. Every fixed amount of radians or time, the cycle repeats. This is seen if the graph is plotted.
  4. May 31, 2010 #3
    Umm, well the way they get the equation. The text says "we can derive a formula for the period of simple harmonic motion by comparing it to an object rotating in a circle"

    How they get their variables, using the object rotating in a circle, is what I find confusing. For example: why choose "sin" when the object is at the equilibrium position (and t = 0)?
  5. May 31, 2010 #4


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    If you inspect the graph of x=sint, you will see that at x=0, t= 0. Meaning that they are starting off with the object not moving from it's equilibrium position.

    Had the stretched it to its maximum value, they'd use x=cos(t).

    So essentially, starting from equilibrium position, use sine.

    Starting from maximum position, use cosine.
  6. May 31, 2010 #5

    I uploaded the picture from my text.

    So what I was wondering, if the object is directly in the middle (x = 0), the "A" line would be straight up and down. Therefore, where does sin come in if there is no "x" to form a triangle like that in picture (a)?
  7. May 31, 2010 #6


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    Normally, you displace the mass at an angle θ from the equilibrium position. This is done so that you can arrive at an equation of motion for the mass. It should be noted that, 'x' is a horizontal displacement, so 'x' will not be a sine but a cosine.


    'A' is a constant radius is in the circle so if viewed from the side, the maximum distance the mass moves on either side of x=0 is A.

    When x=0, you get Acosθ or θ=π/2. So if you view from the top (circle), you will see the mass directly above x=0. If you view from the side (harmonic motion) you will see the mass at x=0.
  8. May 31, 2010 #7
    Ok, but if you start at x = 0 (the equilibrium position) and then push the object, you use sine?

    My book says, "If at t = 0 the object is at the equilibrium position and the oscillations are begun by giving the object a push to the right (+x), the equation would be:

    x = A(sin)wt = A(sin)[2(pi)t/T]"

    I understand the cosine; but just don't know why they use sine?
  9. May 31, 2010 #8


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    Yes will it is just like how your book says, since t=0 should correspond to x=0, the only functions you have to use are sin(t) and cos(t), sin(t) fits this best.
  10. May 31, 2010 #9
    Ok, I think I can do good just memorizing that.

    But Looking at the diagram I attached, if there is no "x" (because the "A" is perpendicular to the x-axis --> "A" is at the 12 o'clock position on the circle) where does sine come from? There should be no angle because there's no x value?
  11. May 31, 2010 #10


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    In your diagram, you are measuring θ anti-clockwise, so at that position, x=0 when θ=π/2 radians or 90°.

    But given how that diagram is drawn with the arrows as is, it looks like they started at x=A and then displaced (since the angle is drawn how it is), so for that motion, the equation would be x=Acosθ or x=Acosωt.

    If they wanted to show the correct diagram for the sine motion, what they should have done was draw the angle measured from the 12 o'clock position. (as that would show x=Asinωt). OR they should have drawn the observer at the adjacent side of the table.
  12. Jun 1, 2010 #11
    Ok I think I got it. Thanks for your help!
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