Period of f(t) = cos(3\pi t) + \frac{1}{2}sin(4\pi t) | T Calculation

[ChickenChakuro]

Homework Statement

What is the period T of $$f(t) = cos(3\pi t) + \frac{1}{2}sin(4\pi t)$$

So, I think I have to find the LCD of 3pi and 4pi, which is 12pi. I don't think this is even close to correct though!

 

[sutupidmath]

well here is the idea on how to show whether a function is periodic or not, but i cannot bother to look to your function, i am just too tired.
If a function is periodic then f(T+t)=f(t), where T is the period of that function. so you need to try to show that

cos[3*pi*(T+t)]+1/2sin[4*pi(T+t)]=cos(3pi t)+1/2 sin(4 pi t),
see if you can get fomething from this!

 

[ChickenChakuro]

While I know this is the definition of a period, I'm not really getting it -- any other hints? Thanks.

 

[sutupidmath]

try to use the cos(x+y)=cosxcosy-sinxsiny, and also the double angle formula for sin and cos, and see if you can cancle something out on both sides, and come up with something like this one one side

cos(..T...)sin(..T...)=0, so from here now you can let either cos(.T...)=0 or sin(..T.)=0
and determine for what value this is true, but you must have a T somewhere either in the sine or cosine or both of them. All you need to do to get there is to apply some trig identities.

 

[sutupidmath]

Or maybe it is better to use these formulas

sinx-+siny=...
and

cosx-+cosy=...
Try both.

 

[Tedjn]

You know that both sine and cosine have period $2\pi$. In your specific function, $\cos(3\pi t)$ and $\sin(4\pi t)$ have periods less than $2\pi$. What are they? If you know that, then find the least common multiple of the two. Use that to find the period of the whole function.

 

[ChickenChakuro]

So, the periods of the cos and sin are 3/2 and 2, respectively? I don't know if this makes any sense...

 

[Tedjn]

They should be less than $2\pi$ because the coefficient in front of t is greater than 1.

 

[ChickenChakuro]

Oh, right, sorry, so the periods of cos and sin are 2/3 and 1/2, respectively, so the LCM of the two is 3?

 

[Rainbow Child]

The LCM of 2/3 and 1/2 is 2.

 

[Tedjn]

I think it is 3. You therefore know that if your cosine portion and your sine portion started off at a certain value each, both will be back at that certain value after t changes by 3. I don't think that constitutes a proof that 3 is the smallest number that would give this function this property, i.e. the period, but it is a good step forward at least.

 

[Rainbow Child]

I think it is 3. You therefore know that if your cosine portion and your sine portion started off at a certain value each, both will be back at that certain value after t changes by 3.

No! If $t$ changes by 3 then you have

$$f(t+3)=\cos3\,\pi\,(t+3) + \frac{1}{2}\,\sin4\,\pi\, (t+3)=\cos(3\,\pi\,t+9\,\pi)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)=-\cos(3\,\pi\,t)+\frac{1}{2}\,\sin(4\,\pi\,t+12\,\pi)\neq f(t)$$

so the period is not 3.

In order to find the period of the above function you have to work like this.
The period of the functions $h(x)=\sin\alpha\,x,\,g(x)=\cos\alpha\,x$ is $T=\frac{2\,\pi}{|\alpha|}$. Of course every multiple of $T$ is a period too, in the sense that the functions repeat their values after $n\,T,\,n\in \mathbb{N^*}$. For the function $f$ the period of cosine is $T_1=\frac{2}{3}\quad\text{or}\quad\tau_1=\,n\frac{2}{3}$ and for the sine $T_2=\frac{1}{2}\quad\text{or}\quad\tau_2=m\frac{1}{2}$. Now in order for the two functions to have the same period, their periods must be equal, thus

$$n\frac{2}{3}=m\frac{1}{2}\Rightarrow n=\frac{3}{4}\,m$$

Thus the first integer value of $m$ which makes $n$ also an integer is $m=4\Rightarrow n=3$ and the period of $f$ is $T=2$.

 

[Tedjn]

Oh, that's right. I stand corrected. For some reason, when adding in my head, I managed to get 3 in my head from 2/3, but that's clearly wrong now. Don't know what I was thinking, so thanks for the catch.