Simple Harmonic Motion
[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? 
Re: Simple Harmonic Motion
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)

Re: Simple Harmonic Motion
Quote:

Re: Simple Harmonic Motion
Quote:

Re: Simple Harmonic Motion
Quote:

Re: Simple Harmonic Motion
1 Attachment(s)
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). 
Re: Simple Harmonic Motion
Quote:
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. 
Re: Simple Harmonic Motion
Quote:
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] 
Re: Simple Harmonic Motion
Quote:

Re: Simple Harmonic Motion
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] 
All times are GMT 5. The time now is 12:37 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums