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Simple Harmonic Motion

  1. Jan 24, 2013 #1

    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?
  2. jcsd
  3. Jan 24, 2013 #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)
  4. Jan 24, 2013 #3
    Hmmm... That seems weird.
  5. Jan 25, 2013 #4
    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.
  6. Jan 25, 2013 #5


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    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.
  7. Jan 27, 2013 #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:
    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 Files:

  8. Jan 28, 2013 #7
    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.
    Last edited: Jan 28, 2013
  9. Jan 28, 2013 #8
    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
    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]
    Last edited: Jan 28, 2013
  10. Jan 29, 2013 #9
    Then how is arcsin a function? I think I get it now. Dang. My trig is rusty.
  11. Jan 30, 2013 #10


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    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]
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