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Deriving angular frequency for simple harmonic motion

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Derive the equation for angular frequency for simple harmonic motion of a spring.

    2. Relevant equations
    Derive omega = sqrt(k/m) from F = -kx
    (sorry i don't know how to use notation)

    3. The attempt at a solution
    I asked my teacher how to do this, and he used some crazy math I didn't learn yet, including Euler's identity and differential equations. I'm in an AP calculus bc class, and i understand differential equations, just not some aspects. Does anyone know a simple solution for this? Thanks in advance
  2. jcsd
  3. Mar 25, 2008 #2
    If [tex]F=ma[/tex] and [tex]F=-kx[/tex]

    Then [tex]ma=-kx[/tex] (by equating the forces.)

    Which can be also written as [tex]ma+kx=0[/tex]

    or [tex]a+\frac{k}{m}x=0[/tex]

    Now if x is displacement, differentiating once with respect to time will give you velocity of the spring and then differentiating again with respect to time will give acceleration.

    Displacement of a spring can be given by

    [tex]x=A * Cos (\omega t)[/tex]

    where A is the Amplitude of motion and [tex]\omega [/tex] is the angular frequency

    Now Differenting once will give velocity;

    [tex]v=-A\omega Sin(\omega t)[/tex]

    and again to give acceleration

    [tex]a=-A \omega^{2} Cos(\omega t)[/tex]

    Now substituting our formula for Acceleration and displacement into our equation of motion


    Gives [tex]-A \omega^{2} Cos(\omega t) +\frac{k}{m}A Cos (\omega t)=0[/tex]

    Which can be rearranged to;

    [tex]A(-\omega^{2} +\frac{k}{m})Cos(\omega t)=0[/tex]

    Can get rid of the [tex]A[/tex] and [tex]Cos(\omega t)[/tex]

    which leaves [tex]-\omega^{2} +\frac{k}{m}=0[/tex]

    which can be rearranged to [tex]\omega=\sqrt{\frac{k}{m}}[/tex]
  4. Mar 25, 2008 #3


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    Homework Helper

    or from here

    that is in the form [itex]a=-\omega^2x[/itex]

    where [itex]\omega^2=\frac{k}{m}[/itex]
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