Understanding Simple Harmonic Motion: Position and Velocity Functions Explained

In summary, the arcsin and arccos functions have a restricted interval so that you only get one solution otherwise you can't even define them. The arctan function has a wider range.
  • #1
tahayassen
270
1
[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?
 
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  • #2
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
0xDEADBEEF said:
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.
 
  • #4
tahayassen said:
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.
 
  • #5
mickybob said:
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.
 
  • #6
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).
 

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  • #7
WannabeNewton said:
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.
 
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  • #8
tahayassen said:
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]
 
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  • #9
mickybob said:
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.
 
  • #10
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]
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where an object oscillates back and forth around an equilibrium position, with a restoring force that is directly proportional to the displacement from the equilibrium position. This means that the further the object is from its equilibrium position, the stronger the force pulling it back towards that position.

What is the position function in Simple Harmonic Motion?

The position function in Simple Harmonic Motion describes the displacement of the object from its equilibrium position at any given time. It can be represented by a sine or cosine function, with the amplitude representing the maximum displacement and the period representing the time it takes for the object to complete one full oscillation.

What is the velocity function in Simple Harmonic Motion?

The velocity function in Simple Harmonic Motion describes the rate of change of the displacement at any given time. It can be represented by a cosine or negative sine function, with the amplitude representing the maximum velocity and the period representing the time it takes for the object to complete one full oscillation.

What is the relationship between the position and velocity functions in Simple Harmonic Motion?

The position and velocity functions in Simple Harmonic Motion are related by the derivative and integral operations. The velocity function is the derivative of the position function, meaning it is the rate of change of the position. And the position function is the integral of the velocity function, meaning it is the accumulated change in position over time.

How is Simple Harmonic Motion used in real life?

Simple Harmonic Motion is seen in many real-life applications, such as the motion of a pendulum, a mass-spring system, and the vibration of guitar strings. It is also used in engineering and design to create stable and efficient structures, as well as in the study of waves and oscillations in physics and mathematics.

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