Simple harmonic motion: why cant you divide by cos?

AI Thread Summary
Dividing by cosine in the equation for simple harmonic motion is problematic because cosine can equal zero, leading to undefined results. The discussion highlights that while the displacement function repeats every 2π, dividing by cosine does not yield useful information and can lead to confusion. Angular frequency applies to any periodic motion, not just circular, as it is defined as the product of frequency and 2π. The term "angular" refers to how it relates to phase changes in oscillatory motion. Understanding these concepts clarifies the mistakes made in manipulating the equations.
connor415
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The displacement, x= Acos(\varpit+\phi), repeats for every increase in 2pi. Why can't I make the above equal to Acos(\varpit+\phi + 2pi), and divide by cos. This gives \phi]=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(\varpit+\phi), repeats for every increase in 2pi. Why can't I make the above equal to Acos(\varpit+\phi + 2pi), and divide by cos. This gives \phi]=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

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

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\pi, 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 f\left(x\right) and g\left(x\right):

\frac{f\left(x\right)}{g\left(x\right)} \neq \frac{x}{x}

Specifically,

\frac{\cos\theta}{\cos\phi} \neq \frac{\theta}{\phi}

Or for a numerical example:

\frac{\cos\left(2\pi\right)}{\cos\left(\pi\right)} = \frac{1}{-1} = -1 \neq \frac{2\pi}{\pi} = 2

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
 

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