(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am asked to derive the equation for angular frequency:

[itex]\omega[/itex]=(2[itex]\pi[/itex]/T)

=2[itex]\pi[/itex]f

2. Relevant equations

X(t)=XCos([itex]\omega[/itex]t +[itex]\phi[/itex])

3. The attempt at a solution

Here's my attempt. I am following Halliday Resnick, but I don't understand the logic.

Given that [itex]\phi[/itex]=0, and the cosine function repeats at every 2[itex]\pi[/itex], this part makes sense to me:

XCost[itex]\omega[/itex]t = XCos[itex]\omega[/itex](t + T)

But here is what I don't understand.

Xcos[itex]\omega[/itex]t = XCos([itex]\omega[/itex]t + 2[itex]\pi[/itex]t)

I don't understand how the book goes onto this next step:

[itex]\omega[/itex](t + T)= [itex]\omega[/itex]t + 2[itex]\pi[/itex]

[itex]\omega[/itex]T = 2[itex]\pi[/itex]

???

What happened to the Cosine functions?

After somehow obtaining that relationship, it makes sense to me.

Simply divide by T, and get

[itex]\omega[/itex]= (2[itex]\pi[/itex])/t

And thus, put 1/t=f, so [itex]\omega[/itex]= 2[itex]\pi[/itex]f.

But I don't understand how they equated omega to 2pi.

Also, now that I am asking this question, I might as well ask how the original function X(t)=XCos([itex]\omega[/itex]t + [itex]\phi[/itex]) comes from. The thing about these introductory physics courses is that they do not explain the basic equations, expecting me to know them already.

This was today's lesson, and it was very confusing. Later during the lesson, we somehow equated omega to sqrt(k/m), but that doesn't make sense to me either, considering how I just finished learning that omega= angular velocity. Well, I guess that is another post though.

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# Homework Help: Derivation of angular frequency.

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