Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Having difficulty with this Oscillation problem

  1. Jun 1, 2004 #1
    This question is quite interesting:

    A uniform meter stick of mass M is pivoted on a hinge at one end and held horizontal by a spring with spring constant k attached at the other end. If the stick oscillates up and down slightly, what is it's frequency? The length of the stick is one meter.

    My thoughts: I believe I should write a torque equation about the hinge, but I am having difficulty doing that--- any suggestions?

    Also and I f=[(k/M)^.5]/2pi but the back of the book has 3k where I have k. I don't understand how they get this. Thanks a lot for your help!

    Also if you need a picture ill be happy to draw one out --- thanks again.
     
  2. jcsd
  3. Jun 1, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Clearly, as the rod rotates an angle [tex]\phi[/tex] about the hinge, there is a slight horizontal displacement of the rod (and spring) as well.
    In the following, I neglect this displacement component, and model the problem with a pure vertical displacement.
    If the displacement angle is [tex]\phi[/tex] then the associated vertical displacement of the tip of the rod (fixed to the spring) is [tex]L\phi[/tex] where L is the length of the rod (to the order of accuracy indicated).

    The moment of inertia about the hinge must be: [tex]I=\frac{ML^{2}}{12}[/tex]
    We gain thus:
    [tex]kL^{2}\phi+\frac{ML^{2}}{12}\frac{d^{2}\phi}{dt^{2}}=0\to\omega^{2}=12\frac{k}{M}[/tex]

    Since [tex]f=\frac{\omega}{2\pi}[/tex] we have:
    [tex]f=\frac{\sqrt{\frac{3k}{M}}}{\pi}[/tex]

    The factor 2 in the denominator in the book is wrong.
     
  4. Jun 1, 2004 #3
    Ahh thank you much I understand all--- I know it's essential to solve the problem but WHY exactly do we use the [tex]kL^{2}\phi[/tex] ?--- or I should ask--- why is it "squared"? I am always quite good at solving and getting the correct answer, but not understanding why equations work can be quite annoying.

    Thank you.
     
    Last edited: Jun 1, 2004
  5. Jun 2, 2004 #4
    Uh oh i may have found a problem:

    if you're just using (1/12)M*L^2, wouldn't you be saying that the axis of rotation WAS at the hinge which is at the END of the rod because you're using L which is the length of the rod? so wouldn't you use the (1/3)M*L^2? can you explain this one to me, ald.?

    Interesting because if we use 1/3 then we end up getting the answer in the back which COULD be right?

    Thanks!
     
  6. Jun 2, 2004 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    In vector form, the spring force is [tex]\vec{F}=-kL\phi\vec{j}[/tex]
    The arm from the hinge to the tip is, to same order of accuracy: [tex]\vec{r}=L\vec{i}[/tex]

    Hence, we get the torque:[tex]\vec{r}\times\vec{F}=-kL^{2}\phi\vec{k}[/tex]
     
  7. Jun 2, 2004 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You're right! Of course it should be 1/3 rather than 1/12. Sorry for that
     
  8. Jun 2, 2004 #7
    Ahh thanks again! You help people out so much on this site and I do appreciate that.
     
  9. Oct 2, 2004 #8
    I'm solving this exact same problem at the moment. Question: Why is the weight of the meter stick being ignored in the torque equation. Shouldn't the torque equation be:

    [tex]\tau = -\frac{1}{2}LMg - kL^2\phi[/tex]
     
  10. Oct 2, 2004 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    At a glance, I believe you're right.
    Anyways (as I'm sure you're aware of), this wouldn't affect the frequency (which is what they want)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Having difficulty with this Oscillation problem
Loading...