Graphical Derivation of x = Asin(ωt)

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Homework Statement


Deriving the equation for simple harmonic motion, x = Asinωt, graphically.

Homework Equations


ω = 2πf, where f = 1/T

2. The attempt at a solution
Take a sine curve as the simple harmonic motion (displacement, x, on y-axis; time, t, on x-axis), then transform it.

The min/max is the amplitude, so we can stretch the graph to say that x = Asin(t).

However, I can't quite get my head around where the ω comes from - I realize that there is a horizontal stretch which must be somehow related to the time period, but I can't quite see why it is 2πf.
 
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Izero said:
However, I can't quite get my head around where the ω comes from - I realize that there is a horizontal stretch which must be somehow related to the time period, but I can't quite see why it is 2πf.
Imagine a circle with radius 1 (m, km, take whatever dimension you want). Walk around that circle f times. How far have you walked?
 
Svein said:
Imagine a circle with radius 1 (m, km, take whatever dimension you want). Walk around that circle f times. How far have you walked?
2πf units (and therefore 2πf radians covered). I think I understand what angular frequency is; I just don't seem to be able to relate it to the graph/equation.
 
Last edited:
Izero said:
I think I understand what angular frequency is; I just don't seem to be able to relate it to the graph/equation.
Perhaps this would help.
Animation1.gif

SHM as projection of uniform circular motion...
 
Okay, so y = sin(kx) stretches the graph by a factor of (1/k), right? (compresses it by a factor of k).

So stretching it by T would actually be a transformation of x = sin(t/T), which is x = sin(ft).

Then you want to 'undo' the pi-ness of the x-axis to make the units seconds (and not have the pi scale hanging around), so you want to stretch by 1/(2π)? That means the whole transformation would be x = sin(2πft). And then you add on the amplitude: x= Asin(2πft).

Does that make sense at all?
 
Last edited:
Izero said:
Okay, so y = sin(kx) stretches the graph by a factor of (1/k), right? (compresses it by a factor of k).

Why not just introduce ##\omega## at this point and not be bothered with ##\pi##?
 
Mister T said:
Why not just introduce ##\omega## at this point and not be bothered with ##\pi##?
Because I still can't fit ω into it in my head! I was trying to reason it through so that it made intuitive sense to me, and the use of ω straight off just doesn't click!
 
The sine function requires an angle, eg Sin(Θ), ω is not an angle, it is an angular velocity. The angle is (ωt)