Question

What is the period of `f(t)=sin( t / 36 )+ cos( (t)/ 42)`?

Answer 1

`T = 504pi`

Explanation:

First of all we, know that `sin(x)` and `cos(x)` have a period of `2pi`.

From this, we can deduct that `sin(x/k)` has a period of `k*2pi`: you can think that `x/k` is a variable running at `1/k` the speed of `x`. So, for example, `x/2` runs at half the speed of `x`, and it will need `4pi` to have a period, instead of `2pi`.

In your case, `sin(t/36)` will have a period of `72pi`, and `cos(t/42)` will have a period of `84pi`.

Your global function is the sum of two periodic functions. By definition, `f(x)` is periodic with period `T` if `T` is the smallest number such that

`f(x+T) = f(x)`

and in your case, this translates into

`sin(t/36+T)+cos(t/42+T) = sin(t/36)+cos(t/42)`

From here, you can see that the period of `f(x)` can't be `72pi` nor `84pi`, because only one of the two terms will make a whole turn, while the other will assume a different value. And since we need both terms to do a whole turn, we need to take the least common multiple between the two periods:

`lcm(72pi,84pi) = 504pi`

Answer 2

`1512pi`.

Explanation:

The least positive P (if any ) such that f( t + P ) = f( t ) is befittingly

called the period of f(t). For this P, f(t+nP)=f(t), n =+-1,, +-2, +-3, ...#.

For `sin t and cos t, P = 2pi.`

For `sin kt and cos kt, P = 2/kpi.`

Here,

the period for `sin (t/36)` is pi/18# and,

for `cos (t/42)`, it is `pi/21`.

For the given compounded oscillation f(t), the period P should be

such that it is also the period for the separate terms.

This P is given by #P=M(pi/18)=N(pi/21). For M= 42 and N= 36,

`P=1512 pi`

Now, see how it works.

`f(t+1512pi)`

`=sin(t/36+42pi)+cos(t/42+36pi)`

`=sin (t/36)+cos(t/42)`

#=f(t).

If halve P to 761 and this is odd. So, P = 1512 is the least possible

even multiple of `pi`.