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Simple pendulum

  1. Jun 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the period of oscillations of a simple pendulum ( a particle of mass m suspended by a string of length l in a gravitational field) as a function of the amplitude of oscillations.

    2. Relevant equations
    [tex] T(E) = \sqrt(2m) \int^{x_2(E)}_{x_1(E)}\frac{dx}{\sqrt(E-U(x)} [/tex]

    where T is the period of oscillations

    3. The attempt at a solution
    I only need the expression of E(x) and the problem is pretty much solved but I can't figure out why (which is rather embarassing :frown:) the energy of the pendulum

    [tex] E = \frac{1}{2}ml^2{\phi}^2 - mgl\cos{\phi} = -mgl\cos{\phi_o} [/tex]

    where [tex] \phi [/tex] is the angle between the string and the vertical and [tex] \phi_0 [/tex] the maximum value of [tex] \phi [/tex]

    Any hints?

    Last edited: Jun 15, 2008
  2. jcsd
  3. Jun 15, 2008 #2


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    Hi armis! :smile:

    KE + PE = constant. So …

    Hint: at what angle does KE = 0? :smile:
  4. Jun 15, 2008 #3
    Hi Tiny-tim :smile:

    Yeah, I knew that! At least that... :smile:

    KE is equal to zero at maximum value of [tex] \phi [/tex], isn't it? And the velocity is highest at point where [tex] \phi [/tex] is zero thus the kinetic energy has the highest value as well.

    Let's see:

    [tex] x= l\sdot\sin{\phi} [/tex]
    [tex] y= l\sdot\cos{\phi} [/tex]

    let's assume l doesn't change ( metal or something). Then

    [tex] \frac{dx}{dt}= l\sdot\cos{\phi} [/tex]
    [tex] \frac{dy}{dt}=-l\sdot\sin{\phi} [/tex]

    Thus [tex] KE=\frac{ml^2}{2} [/tex] :confused: Which is totally ridicilous... It must depend on time! Where did I go wrong?
  5. Jun 15, 2008 #4
    After staring at my post for a while I realised that [tex]\phi[/tex] actually does depend on time. Thus I forgot the derrivative of [tex]\phi[/tex] ( correct me if my termionology is wrong please)
    Oh, my godness... I am so terrible at this :smile:

    Actually in the scanned copy of the book the answer part is scanned so badly that one can't see the dot sign above [tex]\phi[/tex]. I was more focused on how to get [tex]\phi[/tex] in there...:uhh:I guess I should look the answers less and focus on the problem instead, that's for sure :approve:

    But I still can't get the second part right
    Last edited: Jun 15, 2008
  6. Jun 15, 2008 #5


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    Hi armis! :smile:

    I think you need some sleep … :zzz:

    You don't need derivatives at all …

    lcosφ - lcosφo is the difference in height (for working out the PE). :smile:
  7. Jun 15, 2008 #6
    I certainly do :smile: but I fell in love in physics just yesterday and just can't stop. This book by Landau by the way from which the exercise is taken is brilliant

    Hmm, why? If
    [tex] x=l\sin{\phi} [/tex]
    [tex] y=l\cos{\phi} [/tex]
    [tex] \frac{dx}{dt}=l\sin{\phi}\frac{d\phi}{dt} [/tex]
    [tex] \frac{dy}{dt}=-l\cos{\phi}\frac{d\phi}{dt} [/tex]
    [tex] KE=\frac{m}{2}({\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2)=\frac{m}{2}l^2(\frac{d\phi}{dt})^2 [/tex]
    which is exactly what is given as the answer or I am completely wrong. How can I do that without derivatives? :confused:

    I understand what you mean but I am still confused :blushing:

    By the way in their given formula if I understand correctly the part [tex] -mg\cos{\phi} [/tex] is PE. Now since [tex] \phi [/tex] is the angle between the string and the vertical then once [tex] \phi [/tex] is 0 then the potential energy has the highest value? I just can't get it. Isn't it supposed to be 0 at [tex] \phi=0 [/tex]?

    Last edited: Jun 15, 2008
  8. Jun 16, 2008 #7


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    Hi armis,

    You don't need derivatives to figure out your energy formula because you can use the standard formulas for kinetic and potential energies. What are those formulas?

    When [itex]\phi[/itex] is 0, the magnitude of that term [itex]( -mg\ell \cos{\phi}) [/tex] is a maximum, but since it is negative, the potential energy will have it's smallest (most negative) value of the motion.

    You don't need the potential energy to have a specific value of 0 at any particular point, because all that matters is the difference in energy between two points.
  9. Jun 16, 2008 #8
    Hi Alphysicist :smile:


    [tex] KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2 [/tex]
    where q are the generalised coordinates and s are the degrees of freedom.

    And if the force is conservative as it is here then
    [tex] F = - \nabla PE(q) [/tex] thus in this case [tex] PE=-Fr [/tex]

    That's as standart as I could figure it out :frown:

    As for the difference in PE:

    [tex] - mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} ) [/tex]

    Is this correct?

    Last edited: Jun 16, 2008
  10. Jun 16, 2008 #9


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    That's right; and so by knowing that in this problem only the angular coordinate is changing with time you can just write down the kinetic energy term.

    Yes, so the potential energy at any particular point you want is given by that formula:

    [tex] - mg\triangle{h} = - mg ( l\cos{\phi} - l\cos{\phi_0} ) [/tex]

    and remember that you can add any constant value you like to the potential energy; so to simplify add the constant [itex]-m g l \cos\phi_0[/itex] to that and you'll get the original energy formula.
  11. Jun 16, 2008 #10


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    Hi armis! :smile:

    I know you've got it now, but let's just clarify what you did here:
    As you say, that's the correct formula for the KE.

    (Though you could have used the formula for converting angular velocity to speed: [itex]v = \omega r[/itex].)

    But the formula without derivatives gave you the PE! :smile:
  12. Jun 16, 2008 #11
    One question though. If I choose the generalised coordinate as [tex]\phi[/tex] and l then I will not arrive at the KE formula I did previously. What am I missing?

    That's still a little bit confusing but I am getting there. At least by now I already understand that all what matters is the difference in potential energy but I still didn't figure the physical interpretation of this :redface:

    !!Exactly... :uhh: This is harder than I thought but I have learned a lot so far. Thanks for your help

    By the way I am free to book suggestions on classical mechanics as by now it's rather clear I should read one. Landau is a pretty tough one
    Last edited: Jun 16, 2008
  13. Jun 16, 2008 #12


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    The specific formula you entered
    [tex] KE = \sum_j^s \frac {m}{2} ( \frac {dq_j}{dt} )^2 [/tex]

    where m is the mass, is not true when the generalized coordinate does not have units of length. If you want to use [itex]\phi[/itex] as your coordinate, then you will end up with the rotational kinetic energy formula [itex](1/2) I \dot\phi^2[/itex]. Since l is not changing with time it does not have a kinetic energy term.

    The alternative, which I would do, is to follow tiny-tim's good advice about using [itex]v=r\dot\phi[/itex] in the kinetic energy formula [itex](1/2) m v^2[/itex]

    The potential energy function here is mgh. By adding the constant mentioned earlier, h is the vertical position relative to the center of the circular path (so that h=0 when the string is horizontal).
  14. Jun 16, 2008 #13
    Oh, I see. Thanks a lot!
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