Simple Harmonic Motion


by tahayassen
Tags: harmonic, motion, simple
tahayassen
tahayassen is offline
#1
Jan24-13, 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?
Phys.Org News Partner Physics news on Phys.org
Information storage for the next generation of plastic computers
Scientists capture ultrafast snapshots of light-driven superconductivity
Progress in the fight against quantum dissipation
0xDEADBEEF
0xDEADBEEF is offline
#2
Jan24-13, 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)
tahayassen
tahayassen is offline
#3
Jan24-13, 10:07 PM
P: 273
Quote Quote by 0xDEADBEEF View Post
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
mickybob is offline
#4
Jan25-13, 03:56 AM
P: 34

Simple Harmonic Motion


Quote Quote by tahayassen View Post
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
WannabeNewton is offline
#5
Jan25-13, 08:27 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 4,916
Quote Quote by mickybob View Post
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
tahayassen is offline
#6
Jan27-13, 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).
Attached Thumbnails
Untitled.jpg  
mickybob
mickybob is offline
#7
Jan28-13, 04:50 AM
P: 34
Quote Quote by WannabeNewton View Post
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:

[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.
mickybob
mickybob is offline
#8
Jan28-13, 04:59 AM
P: 34
Quote Quote by tahayassen View Post
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).
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
[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]
tahayassen
tahayassen is offline
#9
Jan29-13, 10:18 AM
P: 273
Quote Quote by mickybob View Post
Mathematically, there are generally two solutions of
[tex]sin(x)=y[/tex]
in the range 0 [itex]\leq[/itex] y < 2[itex]\pi[/itex].
Then how is arcsin a function? I think I get it now. Dang. My trig is rusty.
vanhees71
vanhees71 is offline
#10
Jan30-13, 08:03 AM
Sci Advisor
Thanks
P: 2,132
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]


Register to reply

Related Discussions
Problems of statics, angular motion and simple harmonic motion Engineering, Comp Sci, & Technology Homework 1
What is the connection between simple harmonic motion and pendulum motion? Introductory Physics Homework 7
Pendulum/Simple Harmonic Motion, what is its energy of motion? Introductory Physics Homework 10
Using Simple Harmonic motion and conservation of motion to find maximum velocity Introductory Physics Homework 3
Simple Harmonic Motion- From Uniform Circular Motion Introductory Physics Homework 5