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Spring-mass oscillate in 1-D on frictionless horizontal surface

  1. Dec 7, 2007 #1
    We have a spring-mass system which oscillates in one dimension on a frictionless horizontal surface. We act an external force on the mass. F=kχo sin(ωt). Let x=Asin(ωt+φ),A>0 and ωο is the natural frequency of the system.
    a) What is the phase of the motion of the mass relatively to the phase of the external force when ω<ωο and when ω>ωο?
    b) What is the phase of the velocity relatively to the phase of the external force when ω<ωο and when ω>ωο?



    If ω<ωο the motion of the mass is in phase φ with the external force and the amplitude of the mass oscillation is greater than the amplitude of the wiggling. As the forcing frequency approaches the natural frequency of the oscillator, the response of the mass grows in amplitude. When the forcing is at the resonant frequency, the response is technically infinite.
    When the forcing frequency is greater than the natural frequency, the mass actually moves in the opposite direction of the motion, the response is out of phase with the forcing. The amplitude of the response decreases as the forcing frequency increases above the resonant frequency.

    I cannot think of anything else. Could anybody please tell me if I am close to the right answer?
     
  2. jcsd
  3. Dec 7, 2007 #2

    Ben Niehoff

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    Nope, not quite. Have you done Fourier transforms yet? The answer to this question is easiest to see if you take the Fourier transform and calculate the transfer function of the system. Then you can graph the phase part from [itex]0 < \omega < \infty[/itex]. It helps to use a logarithmic scale on the [itex]\omega[/itex] axis.

    What you should see is that the phase is bounded by [itex]-\frac{\pi}2 < \varphi < \frac{\pi}2[/itex]. That is, at the most, your mass will either lag or lead the forcing function by 90 degrees. [itex]\varphi[/itex] is only exactly zero when [itex]\omega = \omega_0[/itex].

    At least, I know this to be the case when you have dissipation. But you're talking about a frictionless system, so I'm not sure.

    If you're not familiar with Fourier transforms and transfer functions, you will probably just have to solve the system normally, and use some trig identities to get the answer into a form that will allow you to answer the question. I come from an engineering background, so we all study transfer functions; I don't know if they're even covered in most physics courses.
     
  4. Dec 7, 2007 #3

    Ben Niehoff

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    Here is a link on transfer functions: http://en.wikipedia.org/wiki/Transfer_function

    It focuses primarily on Laplace transforms, but Fourier transforms are a special case; scroll down the page. The Fourier transfer function is also called the "frequency response".
     
  5. Dec 7, 2007 #4
    Thank you very much. I will see what I can do and I will come back.
     
  6. Dec 7, 2007 #5

    Ben Niehoff

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    Actually, here is a much simpler answer. We were both wrong. You are right in that the mass is in phase with the driving force for small [itex]\omega[/itex], and that it is 180 degrees out of phase for large [itex]\omega[/itex]. However, the phase lag varies continuously, and for [itex]\omega = \omega_0[/itex], the mass lags the forcing function by 90 degrees:

    http://www.kettering.edu/~drussell/Demos/SHO/mass-force.html

    Edit: Whoops, I'm still slightly off. This is for a damped system. In your undamped system, you were correct at the beginning!
     
  7. Dec 8, 2007 #6
    I think I am lost now. Does antone have a hint, please?
     
  8. Dec 14, 2007 #7
    What is the difference between the phase of the motion and the phase of the velocity?
     
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