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A Simple Harmonic Motion Problem

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img441.imageshack.us/img441/6951/physics.gif [Broken]

    As shown in the figure, one end (O) of a uniform rod of length L and mass M is pivoted to a wall. Two horizontal springs with force constants k1 and k2 are attached to the other end (P) of the rod. When the rod is aligned exactly along the vertical axis, the system is in equilibrium.

    a) Find the period of small oscillations of the rod.
    b) Assume that the initial angle the rod makes with y-axis is a small angle θ and the initial velocity 0. Find the velocity of the end point P of the rod as it passes through the equilibrium position.

    2. Relevant equations

    Moment of inertia of a rod relative to its end: I = 1/3 ML2

    3. The attempt at a solution

    I think I have a solution but it is too long and complex. I doubt it's correct. Any help appreciated.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 25, 2010 #2
    Moment on the rod due to the spring is 2kl^2*theta. Equate it with I(alpha) then u will gte alpha and hence omega. From here u can proceed easily to find the period
     
  4. Jul 25, 2010 #3
    That's what I did. But I added the moment from the mass of the rod too.

    So k1L2θ + k2L2θ + MgLθ/2.

    That's equal to Iα. (α being the second derivative of θ)

    (k1L2θ + k2L2θ + MgLθ/2) / I is therefore w2.

    That's part a. For part b, wmax = wθ and Vmax = wθL.

    [PLAIN]http://img375.imageshack.us/img375/6378/captureoj.jpg [Broken]

    Are my answers correct? Thanks.
     
    Last edited by a moderator: May 4, 2017
  5. Jul 25, 2010 #4
    yes u r correct....
     
  6. Jul 25, 2010 #5
    Thanks a lot Swap.
     
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