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Find the frequency of the oscillation of a horizontal beam.

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data

    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 its frequency? [Hint: Write a torque equation about the hinge.] The length of the beam is 1.25 m.

    2. Relevant equations

    τ = Iα

    Fspring = -kx

    3. The attempt at a solution

    I have found the question online and have linked it to this comment in case someone needs to see a picture of the scenario.

    I started by writing a torque equation, with the beam angled θ above the horizontal and x being the displacement of the spring tip above the horizontal.

    The torques about the pivot on the wall on the stick, using the small angle approximation, are:

    1) The spring exerts a force, -kx = -kLθ , a distance, L, away from the pivot. Thus, the torque exerted by the spring is, for small angles, -kxL2θ.

    2) The center of mass exerts a torque, given by -mgL/2.

    The torque equation is. therefore, ∑τ = Iθ'' = -kL2θ - mgL/2.

    This is where I was stuck. I don't know how to solve differential equations beyond separable ones as of yet. The book I am using shows that the solution to a simple harmonic oscillator's equation of motion is just a sinusoid; however, this does not seem like a simple oscillator as there is gravity in this system, which is odd as this is the simple oscillatory section of my book.

    I tried substituting a function of cosine; however it does not seem to work. If anyone could shed light as to what I did incorrectly or what I am overlooking, it would be greatly appreciated.

    Thank you in advance.
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2015 #2
    I forgot to mention, the problem in the attachment is number 5.
     
  4. Jul 19, 2015 #3

    Nathanael

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    Homework Helper

    Consider the function y=sin(x)+47

    When you differentiate, the constant will just disappear, so the function will be a solution to y''= -(y-47)

    You could do this more generally to find the solution of y'' = -C1y+C2 (which is the form of your equation) but I just wanted to make a point.

    mgL/2 is just a constant, so the solution to your equation will be a sinusoid plus a constant. Adding a constant doesn't affect the frequency of oscillation (it just shifts the equilibrium position which it is oscillating about).
     
  5. Jul 19, 2015 #4
    Ok. Thank you very much. I get it now.
     
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