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Homework Help: Simple Pendulum concept

  1. Oct 8, 2007 #1
    The figure shows a simple pendulum of length L = 54 cm and mass m = 2.4 kg. It's bob is observed to have a speed of vo = 4.8 m/s when the cord makes an angle θo = 24°. What is the speed of the bob when it is in its lowest position?

    I'm having problems with this type of problem, its a new concept. Mind helping.
  2. jcsd
  3. Oct 8, 2007 #2
    What is the general equation for a pendulum? How does the information you are given relate to that equation? What is happening, physically, when the bob is in its lowest position?

    It's completely inaccurate to use it in this case since it only hold for small amplitude oscillations, but that is a fault of the problem, and you will just have to avoid learning inaccurate physics. Just don't believe all pendulums are described by the equation you know.
  4. Oct 8, 2007 #3
    I=ml^2, is that what you mean
  5. Oct 8, 2007 #4
    That would be if you had a stick pendulum that you would use inertia. For the simple case of your problem the eqn is

    [tex]\theta(t) = A cos(\omega t + \phi)[/tex]

    where A is the amplitude, omega is the angular velocity given by sqrt(g/l), and phi is the phase of the pendulum.
  6. Oct 8, 2007 #5
    what do you mean by phase, and how would i calculate amplitude
  7. Oct 8, 2007 #6
    Phase is how much the cosine wave is offset from the origin. So picture the cosine function and then move it to the right by some amount. That would correspond to a negative phase, where the equation would be

    [tex]\theta(t) = A cos(\omega t - dist_{\textrm{right}})[/tex]

    if you had a positive phase shift, it would move the cosine function to the left

    [tex]\theta(t) = A cos(\omega t + dist_{\textrm{left}})[/tex]

    Typically you will find the amplitude and phase from initial conditions, such as the one you have. You know that the velocity at time 0 and displaced to 24 degrees will be 4.8m/s.

    I will show you what the initial (initial meaning time is zero) 24 degree displacement would give you

    [tex] \theta(0) = 24^o[/tex]

    [tex] 24^o = Acos(\phi)[/tex]

    You have one equation, but two constants to solve for. What other initial condition could you use (hint: angular velocity is the time derivative of the angle)?

    Also, make sure you convert the 24º into radians as soon as possible, what I posted should have been in radians even.
  8. Oct 9, 2007 #7
    I think the stuff that is being given to me is a bit ahead of where I am in physics. This question is just an example of potential and kinetic energy. Right now, the only thing I am sure of is there is no potential energy at the lowest point of the pendulum
  9. Oct 9, 2007 #8
    Quick question:

    is the change in height 1.175
  10. Oct 9, 2007 #9
    Ah, okay. Sorry, this is why it is good to show your work, or talk about some ideas before you ask for help.

    The change in height, the magnitude, is l - lcos24º, whatever that figures out to be.

    So you should probably define your zero potential at the lcos24º height or the l height. Then you know the Energy initial will equal energy final.
  11. Oct 9, 2007 #10

    I'm having a problem setting up the equation for potential energy. That's where it all lies.
  12. Oct 10, 2007 #11
    Is the right answers 6.47 m/s
  13. Oct 10, 2007 #12
    You would get, if you defined the zero potential to be at the height lcos24.

    [tex]\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv_f^2 - mg(l - lcos(24))[/tex]

    Solve for vf.
  14. Oct 10, 2007 #13
    4.894 m/s

    That's what i got with that equation.
  15. Oct 10, 2007 #14
    It's right!
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