Period of Trigonometric Function

[JSmith]

Hi!

The question states:If k = 7π, what is the period?

I am working in radians. Now, normally, k values do not have a pi attached. Now, to determine the period of a trigonometric function, you divide 2pi by k. However, if this happens, would the pi's not cancel out, leaving the answer with no pi value attached? I cant see how this is right? Am i supposed to multiply out the 7(3.14), and then have 2pi/7(3.14)?????

 

[Deleted member 4993]

Hi!

The question states:If k = 7π, what is the period?

I am working in radians. Now, normally, k values do not have a pi attached. Now, to determine the period of a trigonometric function, you divide 2pi by k. However, if this happens, would the pi's not cancel out, leaving the answer with no pi value attached? I cant see how this is right? Am i supposed to multiply out the 7(3.14), and then have 2pi/7(3.14)?????

Does not make sense.

Please post the complete problem.

 

[srmichael]

Hi!

The question states:If k = 7π, what is the period?

I am working in radians. Now, normally, k values do not have a pi attached. Now, to determine the period of a trigonometric function, you divide 2pi by k. However, if this happens, would the pi's not cancel out, leaving the answer with no pi value attached? I cant see how this is right? Am i supposed to multiply out the 7(3.14), and then have 2pi/7(3.14)?????

I know it seems weird to not have a pi in the period, but it is not mandatory to have a pi. In your example the period would be 2/7 radians which is about 16.4º.

 

[JSmith]

I posted the question in its entirety. Thank you, I suppose I don't need a pi after all.

 

[MarkFL]

If k is the angular velocity and the function is sinusoidal, i.e, we have:

\(\displaystyle f(t)=A\sin(kt)+B\cos(kt)+C\)

then the period is indeed:

\(\displaystyle T=\dfrac{2\pi}{k}\)

 

[dsk2]

JSmith,

If you really want pi, think of 7 as (7/pi)*pi.

But as MarkFL mentioned, for a sinusoidal function, this is how you find the time period of the wave. The formula T=2*pi/k (or sometimes 2*pi/omega) makes it really easy.

However, when you calculate periods and phase shifts for, say, tan, the period is n*pi where n=1, 2, 3...
So, the general equation for the period would be calculated differently for different trigonometric functions, but I guess that is not your specific concern now.

Cheers,
Sai.