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Angular Frequency, Time, and Angle

  1. Oct 18, 2004 #1
    pi = 3.14159...
    angular frequency = 2(pi)f
    theta [in radians] = 2(pi)f t

    t = theta / 2(pi)f

    For theta = 0, t = 0
    For theta = 2(pi), t = 1/f

    If 0 =< theta <= 2(pi)
    Then 0 =< t <= 1/f

    [I'm not sure if the notation above
    is correct.]

    Is the foregoing true for any frequency?
    For instance, 10^10 radians/s?

  2. jcsd
  3. Oct 18, 2004 #2
    When [tex]\theta=2\pi[/tex], one whole cycle has been completed. By definition, the time taken for a whole cycle is the period [tex]T[/tex]. The relationship between the period and the frequency is [tex]T=1/f[/tex].

    Note that [tex]\theta[/tex] denotes angular displacement. There is no need to constraint the value of [tex]\theta[/tex] to be less than [tex]2\pi[/tex]. When it is larger than [tex]2\pi[/tex], it just simply means that it has completed more than one cycle. For instance, when [tex]\theta=4\pi[/tex], it has completed two cycles and certainly the time taken will be twice of the period, i.e. [tex]2T[/tex].

    Hope that I do answer your question correctly.

    Best regards,
  4. Oct 19, 2004 #3
    Taking this a step further


    Thanks for reviewing the basics about frequency, period, and angle.
    I still need some help with angles larger than 2pi.

    Take the equation for a sinusoidally varying
    electric field:

    E = Eo cos(2pift)

    In the above, t doesn't have to be the period.

    Now, for a hypothetical example, let:

    2pif = 10^10 radians/s
    and t = 1s

    What does cos(2pift) equal?

    My calculator gives an error message. Doesn't
    the angle have to be be between 0 and 2pi for
    the cosine function to work?

    Or, doesn't the product of f and t have to vary
    between zero and 1? (same thing as above)

  5. Oct 19, 2004 #4


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    Homework Helper
    Gold Member
    Dearly Missed

    The reason why your calculator gives you an error message, is that the argument you've given it is larger than the maximum argument of the cosine function your calculator has been built to handle.
    Basically, you're giving it a major headache, and it responds with a grumpy error message.
  6. Oct 19, 2004 #5
    Getting Closer!

    If I understand you correctly, you are saying that an argument of any size
    is ok.

    Isn't there an algorithm that can be done by hand or by computer program that can take an agument of any size (an angle of any size) and reduce it to an angle between 0 and 2pi?

    Recall that for cos(2pift) the argument (the angle) will always be positive. And, that the cosine function must return a value between -1 and 1.

    Given the angle 10^10 radians. Given that there is another angle between
    0 and 2pi that will return the same value for the cosine function, is the
    smaller angle equivalent to the larger angle?

  7. Oct 19, 2004 #6
    Recall that [tex]\cos(\theta)[/tex] is a periodic function of [tex]\theta[/tex]. The period in this case is [tex]2\pi[/tex]. This function will repeat itself in the range of [tex][0,2\pi], [2\pi,4\pi], [4\pi,6\pi][/tex], etc. Therefore, whatever value of [tex]\theta[/tex] you have, you will always find a value of [tex]\theta[/tex] within [tex]2\pi[/tex] so that the value of the function is the same.

    Hope that this clarifies your doubt.

    Best regards,
  8. Oct 20, 2004 #7

    To illustrate more clearly what I'm after I'd like to switch
    from units in radians to units in degrees.

    Given theta = 1085 deg.

    n = 1085/360 = 3.0138889

    phi = 360(n-3) = 5 deg.

    cos(phi) = cos(theta)

    Now, are the angles 5 deg. and 1085 deg. equivalent?

  9. Oct 20, 2004 #8


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    Yes, they are equivalent.
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