Question

For the function given, state the period :
F(t)= 6 sin (3t- pi/6) -1

A. pi/3
B. 2pi^2/3
C. 2pi/3
D.-1

Answer 1

I think its A

Answer 2

show your solution to show that it is A

Answer 3

Or C.

Answer 4

what is a normal period of just
F(t)= sin(t) ?

Answer 5

A period of a periodic function is the smallest \(p\) such that: \[
f(x+p) = f(x)
\]Typically we can make this simpler by letting \(x=0\): \[
f(p) = f(0)
\]There are various properties of periods that we can use.

For example, we know that for a simple function \(\sin(x)\) will have a period of \(2\pi\).

Answer 6

t?

Answer 7

And how do we know that the period will be 2pi? @wio

Answer 8

Because \(2\pi\) is a full revolution, and \(\sin\) ignores revolutions.

Answer 9

ok. well how do I figure this out?

Answer 10

Okay, first consider the inner function inside of \(\sin\) here.
It is \(3t-\pi/6\).

Answer 11

We want to find \(p\), such that: \[
(3t-\pi/6)+2\pi =3(t+p)-\pi/6
\]

Answer 12

To make it more clear, we have \[
g(t)=3t-\pi/6
\]And want to find the period of: \[
6\sin(g(t))-1
\]So we want: \[
g(t+p) =g(t)+2\pi
\]

Answer 13

Hmm, a better way to do this is to understand that if \(f(x)\) has a period of \(p\), then \(f(ax+b)\) has a period of \(p/a\).

Answer 14

The reason for this is because when \(x=p/a\), we have \[f(ax+b)=f(a(p/a)+b) = f(p+b)=f(b)\]

Answer 15

Isn't the period 2pi/3? The period of a sine function is always 2pi/B, and in this case B = 3, so the period would be 2pi/3.