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Homework Help: Harmonic Oscillator

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A body of mass 4[kgr] is moving along the x axis while the following force is applied on it:

    [tex] F= -3(x-6) [/tex]

    We know that at time t=0 the kinetic energy is [tex] K=2.16[J] [/tex] and that its decreasing, that is, [tex] \frac{dK}{dt}<0 [/tex].
    The potential energy (with respect to the equilibrium point) at that same instant is: [tex] V=0.96[J] [/tex]

    When will this body return to its initial point (i.e. the position at t=0)?

    2. The attempt at a solution

    The initial position of the body with respect to the equilibrium point (x=6) is found from the P.E. expression, as the "spring" constant is k=3. So:

    [tex]V=0.5kx^{2} \to x= \pm 0.8 [/tex]

    Now, we find the position of the body with respect to time. The amplitude is found from energy conservation between the initial state and when the body is at a maximum distance from equilibrium (K=0), so that:

    [tex] K(t=0)+V(t=0) = 0.5kA^{2} \to A=1.442 [/tex]


    [tex] x(t) = 1.442cos(\omega t + \phi) [/tex] where [tex] \omega^{2}=k/m=0.75 [/tex] and [tex]\phi[/tex] is found from the initial position of the body: [tex] x= \pm 0.8 [/tex]

    for the +0.8 we get after substitution: [tex] \phi = 56.3^{o} [/tex]

    for the -0.8 we get after substitution: [tex] \phi = 33.7^{o} [/tex]

    The data of [tex] \frac{dK}{dt} < 0 [/tex] means that the body is drawing away from the initial position, so had it been in x=+0.8 then it moves to the right and had it been in in x=-0.8 then it moves to the left.

    So I have everthing I need to know about this oscillator. My question: What is the condition I need in order to answer what is required, i.e., the condition of the return of the body to its initial place?
    Last edited: Aug 30, 2010
  2. jcsd
  3. Aug 30, 2010 #2


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    First off ω is not k/m. The correct expression is ω2 = k/m. If you find the correct ω the rest is easy because this problem has a lot of red herrings. You are essentially asked to find the period of oscillations. How is that related to ω?
  4. Aug 30, 2010 #3
    It was just a typo :wink: ... the angular frequency is of course: [tex] \omega = \sqrt{3}/2 [/tex]

    and no, it's not the period of oscillations which is [tex]T=2\pi/\omega=7.26 [sec] [/tex] because the correct answer is:

    [tex] t = 2.27[sec] [/tex]
  5. Aug 30, 2010 #4


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    You're being a bit sloppy with your notation, using x to stand for two different quantities. You should have really solved


    and your solution x(t) should be

    [tex]x(t)=6+1.442\cos(\omega t+\phi)[/tex]

    How exactly was the question worded? Does it essentially ask for the period or does it ask for t when x(t)=x(0)? It'll pass through the same point but going in the opposite direction on the way back to the equilibrium point. If it's the latter interpretation, it might help you to look at a graph of x(t) to see how to go about solving the problem.
  6. Aug 30, 2010 #5
    I took my "zero" to be at the equilibrium point so I don't have to account for that "6", as far as I know.

    I haven't cut off anything from the original wording: How much time will elapse until the body returns back to its initial position (the position at t=0)?
  7. Aug 30, 2010 #6


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    Yeah, I know what you were doing and that you understand what's going on. It's just a matter of expressing yourself precisely. You're given what x represents when the problem tells you F=-3(x-6). If you want to talk about the deviation from x=6, you should use a different variable, like δx.
    I'll just repeat my suggestion you plot δx as a function of time and see where it's equal to 0.8. That should suggest to you how to solve the problem if you don't already see it.
  8. Aug 30, 2010 #7
    Well yeah, I have just plotted [rex]\delta x(t) [/tex] in matlab and got this answer by eveluating the relavent points in the graph, however, this is an exam question and I don't think computers were allowed to be used :wink: ...

    Even when I equate [tex] x(t=0)=x(t) [/tex] I don't get the correct time :/
  9. Aug 30, 2010 #8


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    Can't say much without seeing your actual calculation.
  10. Aug 31, 2010 #9
    What value are you using for the initial phase? It is not zero here.
    With the correct value of the phase, you'll get the time without plotting software.
    I've got 2.27 s, by using only a pocket calculator.
  11. Aug 31, 2010 #10
    I took it 56.3 degrees.

    [tex]x(0)=x(t) \to 1.442cos(56.3)=1.442cos((\sqrt{3}/2)t+56.3) [/tex]

    but this doesn't give me 2.27[sec] ...
  12. Aug 31, 2010 #11


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    How exactly did you calculate the time? Don't describe or imply what you did in words. Show the exact calculation.
  13. Aug 31, 2010 #12
    I should convert this phase angle to radian [tex]\to 56.3^{o}=0.983[rad] [/tex]

    [tex]x(0)=x(t) \to 1.442cos(0.983)=1.442cos((\sqrt{3}/2)t+0.983) \to cos(0.983)=cos((\sqrt{3}/2)t+0.983) \to 0.983=\pm ((\sqrt{3}/2)t+0.983)[/tex]

    we ignore the "+" possibility because it will give the initial time, i.e. t=0. So:

    [tex] 0.983=-\frac{\sqrt{3}}{2}t-0.983 \to 1.966=-\frac{\sqrt{3}}{2}t \to t=\left | \frac{1.966*2}{-\sqrt{3}} \right |=2.27 [sec] [/tex]

    Well, it appears that I've made it :hurray:
  14. Aug 31, 2010 #13
    by the way, is this solution what you meant or were you thinking of something else?
  15. Aug 31, 2010 #14


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    You're close, but you got the sign on the phase angle wrong. I plotted your expression, and you can see that it has the body moving back toward the equilibrium point so that dK/dt>0, not dK/dt<0 as given. And x(t=2.27) is clearly not equal to x(t=0).

    The fact you arbitrarily had to throw in an absolute value to get the correct answer should have tipped you off as well that something was amiss.

    Attached Files:

  16. Aug 31, 2010 #15
    Got ya! Thanks alot :wink:
  17. Sep 1, 2010 #16
    The sign of the phase should be - .
    Here is where you use the fact that KE is decreasing at t=0.

    Sorry, I did not see the last post.
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