Simple harmonic motion: why cant you divide by cos?

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connor415
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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 doesn't move in a circle.
 
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no sorry it gives 2pi=0.
 
no sorry it gives 2pi=0.
 
connor415 said:
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?
connor415 said:
no sorry it gives 2pi=0.
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.
connor415 said:
And how does angular frequency apply to a mass on a spring anyways, it doesn't move in a circle.
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.
 
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 don't understand why the fact that it is sometimes zero means that I can't divide by it?

I understand that angular frequency=2pi times f. Why is it called angular then? Is it just a theoretical quantity?
 
connor415 said:
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.
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?
connor415 said:
My mistake is dividing by cos. I don't understand why the fact that it is sometimes zero means that I can't divide by it?
What is one divided by zero?
connor415 said:
I understand that angular frequency=2pi times f. Why is it called angular then? Is it just a theoretical quantity?
Angular frequency is no more a theoretical quantity than frequency. Angular frequency is so called because gives the frequency with which phase changes.
 
ah ok. I see my schoolboy errors now. Been a while ha