Angular Frequency, Time, and Angle

1. Oct 18, 2004

Larry717

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?

Larry

2. Oct 18, 2004

kenhcm

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. $$2T$$.

Best regards,
Kenneth

3. Oct 19, 2004

Larry717

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:

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

4. Oct 19, 2004

arildno

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.

5. Oct 19, 2004

Larry717

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?

Larry

6. Oct 19, 2004

kenhcm

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.

Best regards,
Kenneth

7. Oct 20, 2004

Larry717

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

8. Oct 20, 2004

Pyrrhus

Yes, they are equivalent.