Derivation of angular frequency.

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SUMMARY

The discussion focuses on the derivation of the angular frequency equation, specifically \(\omega = \frac{2\pi}{T} = 2\pi f\), where \(T\) is the period of oscillation and \(f\) is the frequency. The relationship is established through the periodic nature of the cosine function, which repeats every \(2\pi\). The confusion arises from understanding how the cosine function relates to the angular frequency and the phase shift represented by \(\phi\). The discussion also touches on the connection between angular frequency and the formula \(\omega = \sqrt{\frac{k}{m}}\) in the context of simple harmonic motion (SHM).

PREREQUISITES
  • Understanding of angular frequency and its relation to oscillation
  • Familiarity with the cosine function and its periodic properties
  • Basic knowledge of simple harmonic motion (SHM)
  • Concept of phase shift in wave functions
NEXT STEPS
  • Study the derivation of the simple harmonic motion equations
  • Learn about the properties of the cosine function in relation to periodic functions
  • Explore the concept of phase shifts in wave mechanics
  • Investigate the relationship between angular frequency and spring constants in oscillatory systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and wave motion, as well as educators looking for clarification on angular frequency and its applications in simple harmonic motion.

Scintillation
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Homework Statement


I am asked to derive the equation for angular frequency:
\omega=(2\pi/T)
=2\pif

Homework Equations


X(t)=XCos(\omegat +\phi)


The Attempt at a Solution


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

Given that \phi=0, and the cosine function repeats at every 2\pi, this part makes sense to me:
XCost\omegat = XCos\omega(t + T)

But here is what I don't understand.
Xcos\omegat = XCos(\omegat + 2\pit)

I don't understand how the book goes onto this next step:
\omega(t + T)= \omegat + 2\pi
\omegaT = 2\pi

?
What happened to the Cosine functions?

After somehow obtaining that relationship, it makes sense to me.
Simply divide by T, and get
\omega= (2\pi)/t
And thus, put 1/t=f, so \omega= 2\pif.

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(\omegat + \phi) 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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Scintillation said:

Homework Statement


I am asked to derive the equation for angular frequency:
\omega=(2\pi/T)
=2\pif

Homework Equations


X(t)=XCos(\omegat +\phi)

The Attempt at a Solution


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

Given that \phi=0, and the cosine function repeats at every 2\pi, this part makes sense to me:
XCost\omegat = XCos\omega(t + T)

But here is what I don't understand.
Xcos\omegat = XCos(\omegat + 2\pit)

I don't understand how the book goes onto this next step:
\omega(t + T)= \omegat + 2\pi
\omegaT = 2\pi

?
What happened to the Cosine functions?

After somehow obtaining that relationship, it makes sense to me.
Simply divide by T, and get
\omega= (2\pi)/t
And thus, put 1/t=f, so \omega= 2\pif.

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

T is the period of the oscillation, and since the oscillation is described by a cosine function, it's also the period of the function.

Defining things fairly loosely, the period of a function f(x) is the *smallest* positive amount (call it 'a') you need to add to the argument (x, in this case), so that the value of the function remains the same. In other words f(x+a) = f(x), no matter what the value of x.

Note the qualifier "smallest positive amount". This is important. Because f(x) = f(x+a) = f(x+2a) = ...f(x+na), where n is a positive integer. But the period remains a because you're looking for the smallest positive amount.

Now when you say the period of the cosine function is 2∏, it means cos(x) = cos (x + 2∏). Of course, just like above, cos (x + 4∏), cos (x+6∏) etc. will also give you the same value, but the period is still 2∏, for the reason discussed above.

So the oscillation is defined by a time-variable function. x(t) = x_0\cos{\omega}t is the usual form. This means that x (which usually represents displacement from the equilibrium position, generally defined as zero) starts at its maximum value of x_0 initially (when time t = 0), then goes down to 0, then goes further down to its minimum of -1, then swings back upward. Since we've *defined* the period of the oscillation to be T, we can write: x(t+T) = x_0\cos{\omega}(t + T) = x(t). This means that after time T, the function returns to the same value that it had at time t. In physical terms, this means the oscillator returns to the same displacement x(t) that it had at time t.

But you also know, from the math, that the cosine function has an angular period of 2∏. You can therefore write: x_0\cos{\omega}(t + T) = x_0\cos({\omega}t + 2\pi).

You also know that since the period is always defined by the *smallest* amount that you need to add to the argument of the function, the two quantities are in fact equal. That's how you justify:

{\omega}(t+T) = ({\omega}t + 2\pi)

{\omega}T = 2\pi

T = \frac{2\pi}{\omega}

and since you know (by definition) that T = 1/f,

f = \frac{\omega}{2\pi}

and \omega = 2{\pi}f

Hope that clarifies.

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

The extra \phi is called a "phase". This is an offset added to the function to shift it horizontally along the time axis. This means that instead of having a value of x_0 at time 0, the function will now have a value of x_0\cos\phi. That's all it means. The shape of the function, the period, maximum amplitude, etc. all remain the same. Phase differences become important when you have to model physical processes that can cause phase shifts - e.g certain optical phenomena when you model light waves, inductors and capacitors when you model alternating current circuits, etc.

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.

That's just using the basic definitions to describe a spring-based oscillator. k usually refers to the spring stiffness constant, and m refers to the mass attached to the spring. If you study the basic equations of SHM (Simple Harmonic Motion), you can see the derivation of this result quite easily.
 
Last edited:

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