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X = Asin(wt + angle) and circular motion diagram?

  1. Nov 20, 2013 #1
    This is the type of diagram I'm talking about:


    It's the image the image next to Quest 3T

    If x = Asin(wt) surely the x value is the length of the opposite side, not the displacement of the object in SHM?
    I understand if x = Acos(wt) but why is it always written x = Asin(wt)?!
    sin(wt) doesn't give that length along the horizontal axis?!!!
  2. jcsd
  3. Nov 20, 2013 #2
    I really don't understand what you're asking, your link didn't really work. Maybe you could explain your question better?

    The general solution to the SHM equation is ##x(t)= A\cos (\omega t+\phi) + B\sin (\omega t+\phi)##.

    The equations that link was using were from using a trig identity on the general solution. The idienty: ##\cos (C+D) = \cos (C)\cos (D) - \sin (C) \sin (D)##, if you plug in the right values, you end up with the desired equations.

    The key is that the phase constants will be different for the ##\sin## and ##\cos## versions. If you recall that ##\sin (\alpha + \frac{\pi}{2}) = \cos (\alpha )##, it's only a matter of "lining" them up to get an equivalent expression. All of the different versions of the SHM solution have constants that are related by fixed equations.
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