Determining the Period of sin(ax)*cos(bx)

[deusy]

Homework Statement

I'm trying to work out the period of a function of the form sin(ax)*cos(bx), where a =/= b.

I'm trying solve how the values for a and b relate to the period. I've graphed a lot of functions, but I'm struggling to notice any patterns. The period always seems somewhat related to a and b, but I can't for the life of me work out how.

I'd like to be able to show this algebraically for all values of a and b.

Homework Equations

Possibly the double angle identities:
sin2x = 2sinxcosx
cos2x = cos²x–sin²x = 2cos²x–1 = 1–2sin²x

The Attempt at a Solution

I've been trying to take out common values and use the double angle identities, but so far nothing seems to make sense. Am I just missing something obvious?

 

[BvU]

Hello deusy and welcome to PF. Nice mug shot :)

You have something really good to start with, right? You know the periodicty of $\sin x$ is $2\pi$ and for $\cos x$ it is also $2\pi$.
So $\sin (x+P) = \sin(x)$ with $P = 2\pi$. And $\sin (x+nP) = \sin(x)$ with $n$ an integer number.

Now can you figure out the periodicity for $\sin(ax)$ ? Definitely has something to do with a :)
Same question for $\cos(bx)$.

If you have two periods (different because they tell us $a \ne b\;$), then what is the period of the product ?
Think of what conditions it must fulfil: it has to be an integer multiple of .., and also ..

 

[Chestermiller]

You can't generally be sure that your function is periodic. But you can always express it as the sum of two periodic functions, with generally different periods.

 

[deusy]

If you have two periods (different because they tell us a≠ba \ne b\, then what is the period of the product ?
Think of what conditions it must fulfil: it has to be an integer multiple of .., and also ..

I know the period is 2pi/a for sin(ax), and the same (2pi/b) for cos(bx).

I'm not sure what you mean about the integer multiple bit? Would you mind explaining a bit more? :)

 

[BvU]

Hello again,

Yes, good. So you know that $\sin\left ( a (x + {2\pi\over a})\right) = \sin(ax)$. If you use this twice, you see that also $\sin\left ( a (x + {4\pi\over a})\right) = \sin(ax)$. So :

If a function repeats with a period of $2\pi/a$, it also repeats after twice that, three times, etcetera. If one of these multiples coincides with a multiple of $2\pi/b$, you have found an interval over which the product repeats !

This only works in cases where the ratio a/b is rational; that's why it would be nice to know if there were limitations to a and b in the original problem description. If there aren't, then stating such conditions will be part of the answer.

The other approach (writing the product as a sum) is also worth investigating. But my feeling is that it doesn't help all that much to make things easier. Chet ?

 

[Chestermiller]

I think both methods lead to the same conclusion.

Chet

 

[deusy]

Hello again,

Yes, good. So you know that $\sin\left ( a (x + {2\pi\over a})\right) = \sin(ax)$. If you use this twice, you see that also $\sin\left ( a (x + {4\pi\over a})\right) = \sin(ax)$. So :

If a function repeats with a period of $2\pi/a$, it also repeats after twice that, three times, etcetera. If one of these multiples coincides with a multiple of $2\pi/b$, you have found an interval over which the product repeats !

This only works in cases where the ratio a/b is rational; that's why it would be nice to know if there were limitations to a and b in the original problem description. If there aren't, then stating such conditions will be part of the answer.

The other approach (writing the product as a sum) is also worth investigating. But my feeling is that it doesn't help all that much to make things easier. Chet ?

Eeugh. I think I know what you're getting at but I can't quite grasp it. If the multiples coincide, how do you determine what the period of the overall sin(ax)*cos(bx) function is?

Also, there's no limitations on the values of a and b. I'm trying to work out the period in general form for ALL values of a and b, if that's even possible.

Cheers.

 

[BvU]

I know the period is 2pi/a for sin(ax), and the same (2pi/b) for cos(bx).

I'm not sure what you mean about the integer multiple bit? Would you mind explaining a bit more? :)

A tried my best, but we can do step by step:
I expect the first paragraph in my post #5 is understood. Right ?
And I lose you when I write "If one of these multiples coincides with a multiple of $\; 2\pi/b\$, you have found an interval over which the product repeats !", or not ? (because you ask "how do you determine the period of the over-all function" -- when that's what I had just written):

Eeugh. I think I know what you're getting at but I can't quite grasp it. If the multiples coincide, how do you determine what the period of the overall sin(ax)*cos(bx) function is?

Also, there's no limitations on the values of a and b. I'm trying to work out the period in general form for ALL values of a and b, if that's even possible.

Cheers.

In case of doubt, work out an example...a=4, b=6 is a simple one.

But if I summarize: you got 2pi/a as period of the sine and 2pi/b for the cosine.
Is it clear that n times 2pi/a (n has to be an integer) is also a period of the sine, since it fulfils $$ \sin\left ( a (x + n\;{2\pi\over a})\right) = \sin(ax)\ ?$$
Likewise m times 2pi/b (m an integer) for the cosine factor ?

Then the next bit is somewhat a spoiler:

So that if P = n x 2pi/a = m x 2pi/b both the sin factor and the cos factor are the same for x+P as for x, which means P is the period of sin(ax) * cos(bx) ?

And it's fine with me that there are no limitations on a and b, but if there are no n and m for which n/a = m/b exactly, then the product is aperiodic.

 

[Aryan.Chaudhary]

I was facing same problem. I was looking for a generalized formula. After scratching my head for some time and found the period =2pi/a*m=2pi/b*n. where m and n are smallest possible integers such that a/b=m/n. For example if a=0.1 and b=0.25 then m=2 and n=5. So for given values of a and b the period will be 40 pi. However I have not plotted any graph. Please plot and check it. The formula should work.

 

[haruspex]

I was facing same problem. I was looking for a generalized formula. After scratching my head for some time and found the period =2pi/a*m=2pi/b*n. where m and n are smallest possible integers such that a/b=m/n. For example if a=0.1 and b=0.25 then m=2 and n=5. So for given values of a and b the period will be 40 pi. However I have not plotted any graph. Please plot and check it. The formula should work.

I think you mean a/b=n/m. Yes, that's the right formula. I don't know how you propose to plot a graph, though.