# Simple Harmonic Motion

by tahayassen
Tags: harmonic, motion, simple
 P: 271 $$x(t)=Acos(ωt+ϕ)\\v(t)=-ωAsin(ωt+ϕ)$$ 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?
 P: 827 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)
P: 271
 Quote by 0xDEADBEEF 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.

P: 34

## Simple Harmonic Motion

 Quote by tahayassen 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.
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Thanks
P: 3,988
 Quote by mickybob 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.
 P: 271 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). Attached Thumbnails
P: 34
 Quote by WannabeNewton 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.
P: 34
 Quote by tahayassen 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 $\phi$
P: 271
 Quote by mickybob 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.
 Sci Advisor Thanks P: 1,754 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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