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

1. Nov 20, 2013

applestrudle

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. Nov 20, 2013

Astrum

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.