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Simple harmonic motion: why cant you divide by cos?

  1. Feb 7, 2009 #1
    The displacement, x= Acos([tex]\varpi[/tex]t+[tex]\phi[/tex]), repeats for every increase in 2pi. Why can't I make the above equal to Acos([tex]\varpi[/tex]t+[tex]\phi[/tex] + 2pi), and divide by cos. This gives [tex]\phi[/tex]]=0. This is clearly wrong. Why?

    And how does angular frequency apply to a mass on a spring anyways, it doesnt move in a circle.
     
  2. jcsd
  3. Feb 7, 2009 #2
    no sorry it gives 2pi=0.
     
  4. Feb 7, 2009 #3
    no sorry it gives 2pi=0.
     
  5. Feb 7, 2009 #4

    Hootenanny

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    Well firstly, it's never a good idea to divide by a trigonometric function (unless you restrict the domain) since you are dividing by a function that is sometimes zero. I don't see why you would want to divide by cosine in any case since doing so would yield

    [tex]\frac{\cos\left(\omega t + \phi\right)}{\cos\left(\omega t + \phi + 2\pi\right)} = 1[/tex]

    Which doesn't help you at all. I think that your getting a little confused with the maths.
    Angular frequency isn't simply restricted to circular motion, one can define an angular frequency for any periodic motion. Angular frequency is defined as the product of the frequency and 2[itex]\pi[/itex], so if a system has a frequency, one can define and angular frequency.
     
  6. Feb 7, 2009 #5
    Thank you very much Hootenanny! Ok well when i divided by cos, I got the top bracket equals the bottom bracket, this simplifies to zero. My mistake is dividing by cos. I dont understand why the fact that it is sometimes zero means that I cant divide by it?

    I understand that angular frequency=2pi times f. Why is it called angular then? Is it just a theoretical quantity?
     
  7. Feb 7, 2009 #6

    Hootenanny

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    That is not true. For example, if we have two functions [itex]f\left(x\right)[/itex] and [itex]g\left(x\right)[/itex]:

    [tex]\frac{f\left(x\right)}{g\left(x\right)} \neq \frac{x}{x}[/tex]

    Specifically,

    [tex]\frac{\cos\theta}{\cos\phi} \neq \frac{\theta}{\phi}[/tex]

    Or for a numerical example:

    [tex]\frac{\cos\left(2\pi\right)}{\cos\left(\pi\right)} = \frac{1}{-1} = -1 \neq \frac{2\pi}{\pi} = 2[/tex]

    Do you see your mistake now?
    What is one divided by zero?
    Angular frequency is no more a theoretical quantity than frequency. Angular frequency is so called because gives the frequency with which phase changes.
     
  8. Feb 7, 2009 #7
    ah ok. I see my schoolboy errors now. Been a while ha
     
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