Graphical Derivation of x = Asin(ωt)

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The discussion focuses on graphically deriving the equation for simple harmonic motion, x = Asin(ωt). It begins by examining a sine curve, where the amplitude represents the maximum displacement. The challenge arises in understanding how angular frequency (ω) relates to the time period and the sine function. The transformation involves stretching the graph to account for the time period, leading to the equation x = Asin(2πft) after considering the relationship between frequency and angular displacement. Ultimately, the conversation highlights the difficulty in intuitively grasping the role of ω in the context of the sine function.
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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?
 
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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)
 
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