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Sinusoidally oscillating spring

  1. Jul 26, 2005 #1
    According to Hooke's law [itex] F = k \delta x [/itex], and the energy stored in a certain configuration is [itex] E = \frac{1}{2} k (\delta x)^2 [/itex]

    But it just so happens that this last formula is only valid if one of the ends of this spring if fixed. I would like to know the following:

    What is the elastic energy stored in a spring with natural lengh 1 meter which is performing a sinusoidal oscillation of amplitude (peak to peak) 2 meters and elastic constant equal do 10 N / m.
    Last edited by a moderator: Jul 27, 2005
  2. jcsd
  3. Jul 26, 2005 #2


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    What makes you think so? I would say it is the formula valid in any circumstance, where [itex]\Delta x[/itex] is the amount by which the spring is stretched past its relaxed lenght.
  4. Jul 27, 2005 #3
    Consider the figure I have shown.
    The lower end undergoes a displacement [itex] dx_1 [/itex] and thereby a longitudnal expansion [itex] dx_1 \cos {\theta _1} [/itex]
    Last edited by a moderator: Jul 27, 2005
  5. Jul 27, 2005 #4
    I have read this in some place. I don't remember well. I shall look up. But I really see some disconfort when dealing with a spring which has deformations highly inhomogeneous as the ones we are likely to observe when we get the rope bouncing up and down in a sinusoidal shape (like those ropes children use to jump. Two of them hold the rope and a third stays in between them jumping as it rotates...).

    Let's think the spring is set horizontally and animated by the second harmonic transverse vibration.

    What would be the elastic energy stored in the spring ?(apart from the kinetic energy, of course).
  6. Jul 27, 2005 #5


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    Again, you have attempted to connect two things that are not meant to go together. And you still haven't been able to make better citations of things you "read".

    "spring which has defomations highly inhomogeneous" usually are NOT Hooke Law springs! Just a simple deformations beyond the elastic limit is enough to make Hooke's law not applicable.

    And I'm puzzled how, in your first posting, that you could separate out the Hooke's Law force and the spring PE, and claim that the last one isn't valid under the case you're describing. Since when can the PE expression becomes invalid while the force remains valid? Did you not know that one can be derived from the other? You make one not valid, you also make the other not valid.

    Last edited: Jul 27, 2005
  7. Jul 27, 2005 #6
    hahahaha....you are following me.
    Sorry. You are completelly right about this point. Force law and energy law are interdependent. But the lack of well defined references is a very bad habit, I agree with you. By the way, I don't mean to be rude with you with my comment on tunneling and interference (although I still think they are deeply correlated). Sorry. I have deep respect for you and for what you do.

    I am referring here to a situation where the spring has an S shape because of its oscillation in a second hamonic having its two ends fixed. May be I will have to consider each small part of the spring an independent spring and work on this. Is that correct ? If so, is there any other road?
  8. Jul 27, 2005 #7
    your comment about deformations beyond the elastic limit was already familiar to me.
  9. Jul 27, 2005 #8


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    No, I'm not. I moderate this section of PF (along with Doc Al and other Mentors), so I pay more attention to it than other parts of PF. If you don't want me to "follow you", you can start posting in, let's say, the Philosophy section, for instance.

    Er... so you switched gears just like that and went to standing waves instead? Oy vey.

  10. Jul 27, 2005 #9
    I must say that your objections egarding the ways I am using to express my ideas are really wellcome. Thank you sincerelly.
  11. May 9, 2007 #10
    I would like to check if my solution is ok. I took the sinusoidal shape of the spring, at rest, to be y = A*sin(kx)*sin(wt). At t=0 there is only elastic potential energy, which is the amount I am searching for. At t = 2Pi/w the string is horizontal and all its energy is in kinetic form. I supose that at this instant, all parts of the string are moving along the y-axis. Thus, by taking the derivative of y with respect to time at t= 2Pi/w, we get v_y(x). the diferential of kinetic energy is dE = (1/2) dm * (v_y(x))^2. Now dm equals lambda * dx. Integration in x, finally, solves the problem.

    Is it ok?

    thank you for the attention

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