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 tahayassen Jan24-13 04:15 PM

Simple Harmonic Motion

$$x(t)=Acos(ωt+ϕ)\\v(t)=-ωAsin(ωt+ϕ)$$

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)

 tahayassen Jan24-13 10:07 PM

Re: Simple Harmonic Motion

Quote:
 Quote by 0xDEADBEEF (Post 4241605) 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)
Hmmm... That seems weird.

 mickybob Jan25-13 03:56 AM

Re: Simple Harmonic Motion

Quote:
 Quote by tahayassen (Post 4241967) Hmmm... That seems weird.
Not really. When you do the inverse cos at some point, how do you know which value to take - there are two solutions in the 0 to 2pi interval.

 WannabeNewton Jan25-13 08:27 AM

Re: Simple Harmonic Motion

Quote:
 Quote by mickybob (Post 4242178) Not really. When you do the inverse cos at some point, how do you know which value to take - there are two solutions in the 0 to 2pi interval.
Arccos has a restricted interval so that you only get one solution otherwise you can't even define Arccos. Inverse maps can only be defined if the original mapping is bijective with respect to the codomain.

 tahayassen Jan27-13 03:42 PM

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 $2πn$ where n is an integer.

For example:
$$sin(x)=1\\ x={π\over 2}+2πn$$
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).

 mickybob Jan28-13 04:50 AM

Re: Simple Harmonic Motion

Quote:
 Quote by WannabeNewton (Post 4242305) Arccos has a restricted interval so that you only get one solution otherwise you can't even define Arccos. Inverse maps can only be defined if the original mapping is bijective with respect to the codomain.
Yes, if you want to be a mathematical pedant, this is true.

Although note I did say inverse cos and not arccos.

More correctly then:

$$x(t)=Acos(ωt+ϕ)$$ has two solutions in the interval 0 < $\phi$ < 2$\pi$

In any case, the important thing is the physics of the problem, not mathematical conventions.

 mickybob Jan28-13 04:59 AM

Re: Simple Harmonic Motion

Quote:
 Quote by tahayassen (Post 4245603) 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 $2πn$ where n is an integer. For example: $$sin(x)=1\\ x={π\over 2}+2πn$$ 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).
If you look at the top graph, you'll see that, inbetween A and B, there are two points where the plotted curve crosses any value on the y axis.

Mathematically, there are generally two solutions of
$$sin(x)=y$$
in the range 0 $\leq$ y < 2$\pi$.

$$sin(x)=1$$ 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 $\phi$

 tahayassen Jan29-13 10:18 AM

Re: Simple Harmonic Motion

Quote:
 Quote by mickybob (Post 4246450) Mathematically, there are generally two solutions of $$sin(x)=y$$ in the range 0 $\leq$ y < 2$\pi$.
Then how is arcsin a function? I think I get it now. Dang. My trig is rusty.

 vanhees71 Jan30-13 08:03 AM

Re: Simple Harmonic Motion

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

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