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Harmonic Oscillator: Impulse needed to counteract energy loss

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data

    The pendulum of a grandfather clock activates an escapement mechanism every time it passes
    through the vertical. The escapement is under tension (provided by a hanging weight) and gives the
    pendulum a small impulse a distance [itex] l [/itex] from the pivot. The energy transferred by this impulse
    compensates for the energy dissipated by friction, so that the pendulum swings with a constant
    amplitude.
    a) What is the impulse needed to sustain the motion of a pendulum of length [itex] L [/itex] and mass [itex] m [/itex], with
    an amplitude of swing [itex] θ_0 [/itex] and quality factor [itex] Q [/itex]? You can assume [itex] Q [/itex] is large and [itex] θ_0 [/itex] is small
    2. Relevant equations

    [itex] \ddot{θ}+2β\dot{θ}+(ω_0)^2θ=0 [/itex] where [itex] θ(t)=exp[-βt][θ_0cos(ω_1t)+\frac{βθ_0}{ω_1}sin(ω_1t)] [/itex] from initial conditions [itex] θ(0)=θ_0 [/itex] and [itex] \dot{θ(0)}=0 [/itex]

    [itex] (ω_0)^2 \equiv \frac{mgL}{I}, I =mL^2, 2β \equiv \frac{b}{I} [/itex]

    3. The attempt at a solution

    Okay so first I calculated the initial energy: [itex] E_i = \frac{1}{2}m(ω_0)^2L^2[θ_0]^2 [/itex]

    Next I calculated the energy at a time [itex] t_0 [/itex] later: [itex] E_f = \frac{1}{2}mL^2[\dot{θ(t_0)}]^2 [/itex]

    Then I took the change in energy: [itex] ΔE_- = E_f-E_i = \frac{1}{2}mL^2[[\dot{θ(t_0)}]^2-(ω_0θ_0)^2] [/itex]

    In order to offset this change in energy I need to add an energy [itex] ΔE_+ [/itex] such that [itex] ΔE_-=ΔE_+ [/itex].

    So I'll let [itex] ΔE_+ = \frac{1}{2}ml^2[Δ\dot{θ}]^2 [/itex]

    Since [itex] ΔE_-=ΔE_+ → \frac{1}{2}mL^2[[\dot{θ(t_0)}]^2-(ω_0θ_0)^2] = \frac{1}{2}ml^2[Δ\dot{θ}]^2 [/itex]

    so [itex] Δ\dot{θ} = \frac{L}{l}\sqrt{[[\dot{θ(t_0)}]^2-(ω_0θ_0)^2]} [/itex]

    Now in order to find the impulse [itex] Δp [/itex]: [itex] \|\vec{ΔL}\|=\|\vec{r}\|\|\vec{Δp}\|(1) = IΔ\dot{θ} → Δp = \frac{mL^2}{l}Δ\dot{θ}[/itex]

    So finally I have [itex] Δp = \frac{mL^3}{l^2}\sqrt{[[\dot{θ(t_0)}]^2-(ω_0θ_0)^2]} [/itex]

    I have a feeling I made a mistake in choosing my [itex]ΔE_+[/itex] any help would be greatly appreciated.
     
    Last edited: Mar 2, 2014
  2. jcsd
  3. Mar 2, 2014 #2

    AlephZero

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    Are you making this too hard, because you haven't used
    You could take that assumption as meaning the motion over one cycle is approximately the same as undamped SHM, and then find the work done by the damping force.

    Your solution looks like it will be exact for any value of Q, if you manage to finish it.
     
  4. Mar 2, 2014 #3

    TSny

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    Check to see if there's a typographical error above. The right hand side does not appear to have the dimensions of energy.

    I'm not sure what time ##t_0## refers to.

    I agree with AlephZero that you might be making it more difficult than necessary.

    It appears that they want the answer expressed in terms of the Q factor. Considering the definition of Q factor, can you express the energy lost per cycle in terms of Q and the initial energy?
     
  5. Mar 2, 2014 #4
    I thought about that, but in the solution that my professor gave it did not use the Q factor at all.
     
  6. Mar 2, 2014 #5
    My apologies it should read for the initial energy [itex] E_i = \frac{1}{2}m(ω_0)^2L^2[θ_0]^2 [/itex]. Also, [itex] t_0 [/itex] is a general time during the oscillation of the pendulum. I'm really concerned as to whether my choice for [itex] ΔE_+ [/itex] was correct because my answer looks quite different from the one my professor got in class. The reason why I want to know is because I wouldn't necessarily think to solve it the exact way my professor did.
     
  7. Mar 2, 2014 #6

    TSny

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    Good.

    If ##t_0## is a general time, then shouldn't the expression for the energy at time ##t_0## include potential energy as well as kinetic energy? If you look at your expression for ##E_f## at time ##t_0##, it appears to include only kinetic energy.

    But I'm not sure how you are going to get to the answer with this approach. Since, the Q factor is assumed given, you should be able to find an expression for the amount of energy that must be transferred to the pendulum each cycle. Then deduce what the impulse should be to provide that energy. (You should consider how many impulses there are each cylce.)

    Note, each time the impulse acts, it can be considered as acting over a very small (almost infinitesimal) time interval at the moment the pendulum is vertical. At least, that's how I interpret the problem.
     
  8. Mar 3, 2014 #7
    Ah yes, I believe that's where I went wrong I should have let [itex] t_0 [/itex] be the time of a half cycle so the final energy contains only potential energy.
     
  9. Mar 3, 2014 #8

    AlephZero

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    When ##\theta## is at its the maximum value, the kinetic energy = 0 so the total energy in the system = the gravitational potential energy of the mass.

    I would use that fact, and then use the definition of Q in the link in post #3.

    If your prof did it the hard way by solving the equation of motion, that's his problem not yours :smile:
     
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