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Homework Help: Derivation of angular frequency.

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    I am asked to derive the equation for angular frequency:

    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.
  2. jcsd
  3. Jan 13, 2012 #2


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    Homework Helper

    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. [itex]x(t) = x_0\cos{\omega}t[/itex] 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 [itex]x_0[/itex] 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: [itex]x(t+T) = x_0\cos{\omega}(t + T) = x(t)[/itex]. 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: [itex]x_0\cos{\omega}(t + T) = x_0\cos({\omega}t + 2\pi)[/itex].

    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:

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

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

    [itex]T = \frac{2\pi}{\omega}[/itex]

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

    [itex]f = \frac{\omega}{2\pi}[/itex]

    and [itex]\omega = 2{\pi}f[/itex]

    Hope that clarifies.

    The extra [itex]\phi[/itex] 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 [itex]x_0[/itex] at time 0, the function will now have a value of [itex]x_0\cos\phi[/itex]. 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.

    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: Jan 13, 2012
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