Angular Frequency, Time, and Angle

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Angular frequency is defined as ω = 2πf, where θ represents angular displacement in radians. The time taken for a complete cycle is the period T, with the relationship T = 1/f. Angles larger than 2π indicate multiple cycles, and the cosine function is periodic, meaning it can accept arguments of any size, returning equivalent values for angles that differ by multiples of 2π. For example, an angle of 1085 degrees is equivalent to 5 degrees, demonstrating that angles can be reduced to a standard range for cosine calculations. Understanding these relationships clarifies the behavior of sinusoidal functions across varying frequencies and angles.
Larry717
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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?

Larry
 
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When \theta=2\pi, one whole cycle has been completed. By definition, the time taken for a whole cycle is the period T. The relationship between the period and the frequency is T=1/f.

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

Hope that I do answer your question correctly.


Kenneth
 
Taking this a step further

Ken,

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)

Larry
 
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.
 
Getting Closer!

arildno said:
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.

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?

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

Hope that this clarifies your doubt.


Kenneth
 
Clarification

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?

Larry
 
Yes, they are equivalent.
 
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