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Periodic Driving Force

  1. Mar 12, 2014 #1

    cpburris

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    Not a homework problem, just a question. What is a periodic driving force, specifically what is periodic about it? Is it the magnitude of the force that is periodic?





    1. The problem statement, all variables and given/known data




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    3. The attempt at a solution
     
  2. jcsd
  3. Mar 12, 2014 #2

    Simon Bridge

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    Usually just the magnitude is periodic, but the direction can change too.

    i.e. if you put a plate of jelly on the table, and tap the plate with a spoon once a second, you are applying a periodic driving force to the jelly. You can also grab the plate and slide it back and forth in some repeating pattern - that would be another periodic driving force.
     
  4. Mar 13, 2014 #3

    cpburris

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    So in my mechanics class we are looking at simple 1-D harmonic oscillators. When we talk about a driving force we write it as Fsub0 times the cos of omega(t). Does that mean the driving force is being constantly applied, with the magnitude of the force changing with time in the domain [-Fsub0,Fsub0]?
     
  5. Mar 13, 2014 #4

    AlephZero

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    Yes, unless you say the force in only applied for some time interval, like ##t > 0## or ##t_1 \le t \le t_2##.
     
  6. Mar 13, 2014 #5

    cpburris

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    can you explain this to me then;
    Given
    w = angular frequency of the driver

    wsub0 = angular frequency of the undamped harmonic oscillator

    Equation of motion:
    ma+m(wsub0^2)x=(Fsub0)cos(wt)

    w << wsub0

    "We might expect the inertial term ma to be negligible compared to the restoring force m(wsub0^2)x."

    How that conclusion is drawn eludes me. It doesn't make sense to me.

    Edit: Exploring this I was able to prove this statement mathematically, but it still doesn't seem an intuitive conclusion.

    They go on to say "it follows that the oscillator will move pretty much in phase with the driver."

    Can't see where they came up with that conclusion either.
     
    Last edited: Mar 13, 2014
  7. Mar 13, 2014 #6

    BvU

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    In isolation this doesn't make much sense. The ##\omega << \omega_0## helps, though.

    You wrote the eqn of motion for a harmonic oscillator without damping.
    So any initial conditions that cause a considerable amplitude can give rise to large acceleration.
    If there is some damping, this will be for a certain time only (transient solution), and ultimately only the driving force determines the motion of the oscillator.

    Without actually solving the thing and dealing with the initial conditions, it is a little difficult to say very much about it.

    Usually the mass is in ##\omega_0## and the equation is written as
    $$\ddot x+\omega_0^2 x = F_0\, \cos(\omega t)$$Any solution to the homogeneous equation
    $$\ddot x+\omega_0^2 x = 0$$ can be added to a solution of the complete equation. And such a solution has two integration constants, e.g. ##x(0)## and ##\dot x(0)##.

    If we follow your account, and let the oscillator move with the driver,
    we could try a solution for the inhomogeneous equation
    that looks like ##x(t)=A \cos(\omega\, t+\phi)##, fill it in and get
    $$A\left (-\omega^2 + \omega_0^2 \right) \cos(\omega\, t+\phi) = F_0 \, \cos(\omega t)$$ Which must be satisfied at all t. From which ##\phi = 0## (more than just in agreement with "pretty much in phase with the driver" -- which gives me the impression some damping is considered to be present...)
    and $$ A = {F_0 \over \omega_0^2-\omega^2}$$

    [edit]saw your edit but was too far along already...
     
  8. Mar 13, 2014 #7

    BvU

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    Since A is an amplitude, the expression holds until ##\omega=\omega_0##. At that frequency the amplitude increases linearly with time. For ##\omega>\omega_0## The expression has a minus sign (so that A>0). Consequence: ##\phi = \pi## meaning driver and oscillator move in opposite directions.
     
  9. Mar 13, 2014 #8

    cpburris

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    Yea thanks for the response, I realized my mistake in excluding the mass term in the expression for the restoring force. Still unsure how phi being zero arises from the condition that w<<wsub0. It makes sense iff you also state that the acceleration is small, but if you don't make that distinction, couldn't phi just as easily be pi, making for a large acceleration?

    Edit: ok so I think I get it. If phi was pi, then the equation for the amplitude would change by a negative sign, resulting in the expression only holding true if w>wsub0
     
    Last edited: Mar 13, 2014
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