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Energy of a Damped Pendulum

  1. Jul 9, 2009 #1
    This may straddle more advanced physics, but I thought it leaned toward introductory.

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

    I have been told to find the net energy of a damped pendulum.

    2. Relevant equations

    Obviously the equation of energy for an undamped pendulum is just:

    E = KE + PE = .5mv^2 + mgh = 0

    I know the equation of angular motion for damped pendulum is:

    ø'' - (g/L)sin(ø) - cø' = 0

    3. The attempt at a solution

    As for the Energy Equation of damped pendulum..I'm not certain. I assume it must be along the lines of

    E = .5mv^2 + mgh - ∫Fds.

    where the damping force is some -cv, or cø'. But unlike a damped harmonic oscillator, we're dealing with two dimensions here and that keeps on confusing me.

    As you can see it's a mess of different variable and I can't quite figure out how to structure a decent equation. To make matters more interesting, after I write the equation I have to take the derivative with respect to time.

    All points in the right direction appreciated.
  2. jcsd
  3. Jul 10, 2009 #2


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    Let's start with some questions to help you clarify the problem for yourself, and for those trying to help you...

    Questions for yourself:
    • Do you think the energy of a damped pendulum will have any additional contributions to those you list for the undamped one (i.e. kinetic and potential)? If so what and why?
    • Where are you defining the zero point of your energy? How do you expect the energy to change as the pendulum swings?
    • The equation you give for the angular motion has some sign errors. To correct, ask yourself: would you expect the gravitational force to act to increase [tex]\phi[/tex] or decrease it? Would you expect the damping force to act with the same sign as the velocity or the opposite?
    • What is your definition of [tex]c[/tex] (it seems to be non-standard)? What are its units?

    Questions to help us give you hints:
    • What do you mean by net energy?
    • Are you interested in a general pendulum, or only one undergoing small angle oscillations (the former is non-trivial!)?
    • Are you considering the under-damped, over-damped, or critically damped case (or all three)?
    • Do you know how to solve 2nd order linear ordinary differential equations?
  4. Jul 10, 2009 #3
    1. Yes, I believe the energy of a damped pendulum will have additional contributions to the total energy of the system. I define cø' as some damping force proportional to angular velocity (ø'). C is simply a constant. I suppose its units would then have to be kg/s. Therefore, this force would do work on the system, in this case taking energy out of the system. Work = Force x Displacement. Displacement would be arc length, or r*ø. Energy = -Work = cø'*rø? Still seems pretty wrong.

    2. I'm not certain. Obviously the energy is never zero for an undamped pendulum. I assume for a damped pendulum the work done by the damping force will eventually take enough energy out of the system to make it practically zero.

    3. My fault:

    Torque Gravity = -mglsinø (restoring torque)
    Torque Damping = -cø' (restoring torque)
    Torque Inertial = ml^2*ø''

    ø'' + (g/L)sin(ø) + cø' = 0

    4. All ready answered (to the best of my ability at least).

    5. General undamped. Not small oscillations. I'm afraid I can't make any neat approximations here.

    6. All three, I suppose. This depends on the damping force, which is as of now undetermined, correct? By keeping the damping force defined by some constant c we determine whether it is under, over, or critically damped, I believe.

    7. I do. Though since I'm not making small angle approximations it will do me little good. I should state that my professor told me I would not be able to solve this problem. I just need to be able to set up the equation and get as far as I can.

  5. Jul 10, 2009 #4


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    OK, so you say that you expect the damping to take energy out of the system so that the energy decreases over time. This is correct.

    But you still say you think there should be some additional energy term that is neither kinetic nor potential, so what form of energy do you think we should be adding in? What part/property of the pendulum accounts for this extra energy term?

    If it helps, try to think of it in reverse. If you encompass the entire system energy should be conserved. Since you agree the pendulum is losing energy, where is the flow of energy --- and importantly which part of the energy `belongs' to the pendulum, and which to other parts of the system?

    While you're thinking about that, let's also get started with the meat of the problem. Can you write down in terms of your problem variable ([tex]\phi[/tex]) what the KE and PE terms will look like, taking care to specify e.g. where you are placing the zero of potential energy.

    My point about [tex]c[/tex] is that you aren't being consistent in the various places you use it. Sometimes it includes units of mass, sometimes it doesn't. In a traditional damped oscillator I believe its usual not to have any mass dependence in [tex]c[/tex] --- which would mean you should have [tex]\frac{c}{m}[/tex] where you have just [tex]c[/tex] in your angular motion equation. But it is a minor issue.
  6. Jul 15, 2009 #5
    Maybe I've been on the wrong track. After all, KE + PE = Total E. So I shouldn't be adding a new term, but subtracting from a current one.

    I'm thinking that this resistive force, air resistance/random frictions in reality but that's unimportant in our theoretical model, must be subtracting from the kinetic energy. H, the height that determines the potential energy, is dependent on the velocity of the pendulum. Therefore as this resistive force works against the velocity of the pendulum, both the kinetic and potential energy (aka the overall energy) decrease.

    (1/2)mv^2 + mgh = E

    h(ø) = Lsinø

    v(ø) = 2πø/t

    (1/2)m(v - cv)^2 + mgLsinø = E ?

    But I need to take the derivative with respect to time....:

    E = (1/2)m(2πø/t - c*2πø/t)^2 + mgLsin(wt) ?

    Getting closer?
    Last edited: Jul 15, 2009
  7. Jul 15, 2009 #6


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    It's not clear to me what your objective is. Are you just trying to get expressions for E(t) and Eʹ(t)? If you were using the small angle approximation, then I'm sure you know that you'd be able to solve the DE and would obtain, instead of just oscillations, oscillations that are modulated by an exponentially decaying envelope. Since the total energy of the system is just the grav. potential energy at the extreme ends of the motion, the total energy is related to the value of ϕ on the envelope, and you can see that it is decaying with time.

    Please clarify your objective/exactly what the problem is asking you to do.
  8. Jul 15, 2009 #7
    Yes, all I am trying to do is develop equations for E(t) and E'(t). I am not able to make small angle approximations, and while I know that I will not actually be able to solve the second order differential equation, I just want to set up the equations correctly and then go as far as I can.
  9. Jul 16, 2009 #8


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    Yes! That's right.

    Again correct. So all you need to do is sum the potential and kinetic energies as before: the functional form of these will include the damping effects.

    Unless you've changed your definition of [tex]\phi[/tex] compared to that used in your equation of motion, that isn't correct. Have another think --- ideally upload a diagram showing how you come to your answer so that I can see if it's just a definition issue...

    Also wrong! (You should be able to tell because it doesn't have the right units). Hint: will the answer be a direct function of [tex]\phi[/tex] or a function of its time derivative?

    Finally, why do you want to take the time derivative of the energy (once you've worked it out correctly)?
  10. Jul 16, 2009 #9
    Ah I'm meeting with my professor at 1:30...I'm running out of time to get this energy equation right.

    1) Awesome.

    2) Sweetness.


    cosø = h/L --> h = L - Lcosø. --> L(1-cosø)

    4) I'm really not certain. It's not x(t) = -Awsin(wt) because this isn't SHM, though it is similar. Plus we are dealing in two axes here. I'm just getting more confused. Velocity is changing as a function of angle because it is max when the angle equals zero, and zero at the highest angle. *sigh* I'm running out of time here.

    E(t) = .5m(??)^2 + mgL(1-coswt)
  11. Jul 16, 2009 #10


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    Your image doesn't show (permissions problem at the server), but the above equation is now correct.

    I think you're overcomplicating the issue. Remember you can't find [tex]\phi(t)[/tex] without solving the full equations (which you can't do in closed form for the large angle case). Thus you are left with simply specifying the velocity as a function of the (unknown) solution. Just as the height is a function of the unknown [tex]\phi(t)[/tex] so you should be able to write the velocity as a function of the (unknown) angular velocity [tex]d\phi/dt[/tex] --- that's all I was asking for!

    Don't start introducing [tex]\omega[/tex] as without solving the equations you don't know what it is! Keep everything in terms of [tex]\phi(t)[/tex] and [tex]d\phi/dt[/tex].
  12. Jul 16, 2009 #11
    Is my problem converting angular velocity into linear velocity? Because dø/dt itself is angular velocity. The only thing I can find is people using conservation of energy to solve for the velocity v. Of course, if I did that my equations would just cancel out.
  13. Jul 16, 2009 #12


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    Yes that's all you need do.
    Yes. To convert to the instantaneous linear velocity you just need to work out how to relate a small change in angle, [tex]d\phi[/tex], to a corresponding small change in distance, [tex]ds[/tex], (hint: small angle geometry) and then use this to calculate the linear velocity [tex]v = ds/dt[/tex].

    If you can't spot how to do it then post what you've tried...
  14. Jul 16, 2009 #13
    All right.

    s = rø
    ds/dt = rdø/dt

    ds = rdø
    dø = (1/r)ds

    v = ds/dt = rdø/dt ?
    Last edited: Jul 16, 2009
  15. Jul 16, 2009 #14


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    Technically this should be [tex]ds=r\;d\phi[/tex] or else [tex]s=r\;\Delta\phi[/tex] for finite angles (in which case [tex]s[/tex] is curved).

    Also note that for you pendulum the radius is fixed [tex]r=L[/tex].

    But otherwise you are correct, [tex]v=L\;d\phi/dt[/tex].

    You should now be able to write down the complete expression for [tex]E(t)[/tex] in terms only of [tex]m,g,L,\phi,\;\text{and}\;d\phi/dt[/tex]. This combined with the equation of motion fully specifies the energy: if you were able to solve the motion equation then you could get the energy by substituting the solution into [tex]E(t)[/tex].

    If you want to post your final expression here for somebody to check, then please do, but it should just be algebraic manipulation from this point.

    Further Work

    Of course in many senses, just writing the equations down is only "legwork". The interesting physics comes from thinking about what the system might do, even if you can't solve it exactly. As cepheid hinted at, one could try and just think about what the total energy does on average. It is clear that it will decay: but what form will this decay take. You should be able to get some handle on its form even without reference to the easier harmonic oscillator problem. If you have time (I know you are on a tight schedule) think about the following:

    What would be a good parameter to use to specify the maximum energy in any given oscillation (the behaviour of which could then be used to consider the average behaviour of the energy)?

    What governs how much energy is taken out of the system in each oscillation, and how does this quantity itself evolve over several oscillations?

    What does this say about the likely form of the decay of the "maximum energy per oscillation"?

    At that point you could even turn this into a third exercise: split the equation of motion into two coupled first order equations, write some code to solve them numerically, and then fit your guess above to the numerical solution...

    In short you can often learn a lot by thinking about systems even when they have no closed form solution...
    Last edited: Jul 16, 2009
  16. Jul 16, 2009 #15
    One last question. So the equation I now have is:

    E(t) = (1/2)m(Ldø/dt)^2 + mgL(1-cosø)

    But this is the equation of energy for a pendulum, not a damped pendulum, right?

    So would the equation of energy for a pendulum where energy is constantly being taken out of the system be:

    E(t) = (1/2)m(Ldø/dt - cdø/dt)^2 + mgL(1-cosø)

    I believe that the dampening is taking energy out of the system specifically by decreasing the velocity of the pendulum. That's why I thought to put it inside the KE part of the energy equation. Or is the dampening found within the change in angular velocity, dø/dt?
  17. Jul 16, 2009 #16


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    It is the energy equation for both types of pendulum! The damping (or indeed driving if there were any) comes in through the form of [tex]\phi(t)[/tex] and [tex]d\phi/dt[/tex]. That is why you need to quote both the above energy equation and the equation of motion to fully specify the energy!

    In the case of a pendulum undergoing small angle oscillations you can actually substitute in the solutions to the standard harmonic oscillator equations to see this for yourself: if you substitute in the solution for the motion [tex]\phi_{\rm HO}(t)[/tex] of a damped harmonic oscillator you would find that, after substitution, the damping coefficient naturally features in the energy equation.

    If you like you can think of the energy equation you have above and the equation of motion as two simultaneous equations: only if you could solve this system (which involves finding the solution to the differential equation posed by the motion equation) could you explicitly write down a single energy equation. But that single equation would then include the damping coefficient.

    No! The damping is already implicitly included in [tex]\phi(t)[/tex] and derivatives, you are trying to add it in twice now (and guessing at its form).

    Let me know if this still doesn't make sense and I'll have another go at trying to explain it...
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