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

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SUMMARY

The discussion centers on the mathematical representation of simple harmonic motion (SHM) using the equation x = A sin(wt + φ) versus x = A cos(wt + φ). Participants clarify that while both forms represent the same motion, they differ in phase constants. The general solution for SHM is expressed as x(t) = A cos(ωt + φ) + B sin(ωt + φ). The confusion arises from the interpretation of the sine function in relation to displacement along the horizontal axis, which is resolved through trigonometric identities.

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  • Understanding of simple harmonic motion (SHM)
  • Familiarity with trigonometric identities
  • Knowledge of phase constants in wave equations
  • Basic calculus for interpreting motion equations
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applestrudle
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This is the type of diagram I'm talking about:

http://www.google.co.uk/url?sa=i&rc...VBtPdh26viVXkWaWL6rTIk7w&ust=1385070490503327

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?!
 
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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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