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Simple Harmonic Motion 
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#1
Jan2413, 04:15 PM

P: 273

[tex]x(t)=Acos(ωt+ϕ)\\v(t)=ωAsin(ωt+ϕ)[/tex]
I think my physics professor said in one of the lectures that: after setting up your position function by finding amplitude, angular speed, and solving for ϕ by setting t=0 and using the x(0) value given in the question, you need to to set t=0 in the velocity function and use the v(0) value to make sure your ϕ value is correct. I'm confused. Why is this necessary? 


#2
Jan2413, 05:13 PM

P: 824

If you have some x value you can cross it on the way up or on the way down, this is why you need to check v(0)



#3
Jan2413, 10:07 PM

P: 273




#4
Jan2513, 03:56 AM

P: 34

Simple Harmonic Motion



#5
Jan2513, 08:27 AM

C. Spirit
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P: 5,447




#6
Jan2713, 03:42 PM

P: 273

Okay, this still doesn't make sense to me.
Attached is a graph of f(x) and f'(x). When you do the inverse sign you get a value plus [itex]2πn[/itex] where n is an integer. For example: [tex]sin(x)=1\\ x={π\over 2}+2πn[/tex] Each x value would be 1 period apart. Therefore, as you can see in the diagram, points A, B, and C are all 1 period apart, and f'(A)=f'(B)=f'(C). 


#7
Jan2813, 04:50 AM

P: 34

Although note I did say inverse cos and not arccos. More correctly then: [tex]x(t)=Acos(ωt+ϕ)[/tex] has two solutions in the interval 0 < [itex]\phi[/itex] < 2[itex]\pi[/itex] In any case, the important thing is the physics of the problem, not mathematical conventions. 


#8
Jan2813, 04:59 AM

P: 34

Mathematically, there are generally two solutions of [tex]sin(x)=y[/tex] in the range 0 [itex]\leq[/itex] y < 2[itex]\pi[/itex]. Your example of [tex]sin(x)=1[/tex] is a special case because it's a turning point. In terms of the physics: The pendulum returns to its initial position twice within the period. The difference is that it is now moving in the opposite direction, so the velocity has changed sign. So we need an initial condition for the velocity as well as the position in order to uniquely define [itex]\phi[/itex] 


#9
Jan2913, 10:18 AM

P: 273




#10
Jan3013, 08:03 AM

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P: 2,351

In real analysis the arcsin function is defined as
[tex]\arcsin:[1,1] \rightarrow [\pi/2,\pi/2],[/tex] arccos as [tex]\arccos:[1,1] \rightarrow [0,\pi],[/tex] and arctan as [tex]\arctan:\mathbb{R} \rightarrow (\pi/2,\pi/2).[/tex] 


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