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Proof dealing w/ Simple Harmonic Oscillations

  1. Jan 10, 2004 #1

    nix

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    Prove that the period of a simple pendulum doing small oscillations is equal to:

    2(py)x(square root of: l/g)

    where py is 3.14..(obviously..lol)... l is length of the string of the pendulum and g is gravity

    Also.... the pendulum is basically just a ball on a string moving from side to side. and the equations we have been given to solve it is the fundamental eq. of an oscillation:

    k = mw^2 = m x(2py^2/T) = m (2(py)f^2)

    where f is the frequency, m is the mass, w is omega (the angle), and k is the spring constant and T is period

    tricky..eh?

    thanks for any help or suggestions..
     
  2. jcsd
  3. Jan 10, 2004 #2

    nix

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    oops i just read the thing i was supposed to read before posting...

    anyways this is what ive done and i dont think it makes sense maybe i should clarify the question with my teacher later..

    m(2(py)^2/T) = m(2(py)f^2)

    T = m(2(py)^2)/m(2(py)f^2)
    T = py/f^2

    f = 1/T

    T = (py)T^2..maybe im using the wrong equation...

    And another thing about the question is that the angle is less than or equal to 5 degrees...the angle btw the original position of the pendulum and the position of the pendulum when its movind to the side.
     
  4. Jan 10, 2004 #3
    First equation has got wrong dimensions whereas second equation is correct.

    Equations which might help u are
    [tex]\omega^2 = \frac{k}{m}[/tex]
    [tex]\omega=2\pi f[/tex]
    [tex]T=\frac{2\pi}{\omega}[/tex]
    [tex]kl=mg[/tex]

    All equations are valid for small angle [tex]\theta\leq 5^0[/tex]
     
    Last edited: Jan 10, 2004
  5. Jan 10, 2004 #4

    nix

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    thanks so much for the help with the equations...i got it! yay!

    but can you explain to me how kl = mg...

    thanks again
     
  6. Jan 10, 2004 #5
    That requires u to know the Force equation
    i.e. [tex]F=-mgsin\theta[/tex]
    for small theta [tex]sin\theta = \theta[/tex]
    therefore the equation is [tex]F=-m\theta = -\frac{mgx}{l}[/tex]
    displacement along the arc=[tex]\theta l[/tex]

    also if u draw a free body diagram fo the forces . u have
    [tex]k(l+\delta x)=mgcos\theta[/tex] for small displacement u can assume [tex]\delta x=0 [/tex] & [tex] cos\theta=1[/tex]
    so u get kl=mg
     
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