Question

What is the period of `f(t)=cos ( ( 5 t ) / 2 )`?

Answer 1

`T =1/f= (2pi)/omega = (4pi)/5`

Explanation:

One way to get the period from a sinusoid is to recall that the argument inside the function is simply the angular frequency, `omega`, multiplied by the time, `t`

`f(t) = cos(omega t)`

which means that for our case

`omega=5/2`

The angular frequency is related to the normal frequency by the following relation:

`omega=2 pi f`

which we can solve for `f` and plug in our value for the angular frequency

`f=omega / (2pi)=5/(4pi)`

The period, `T`, is just the reciprocal of the frequency:

`T =1/f= (4pi)/5`