The Fundamental Period

  • Thread starter darkmagic
  • Start date
  • #1
164
0

Homework Statement



The function is defined on (-infinity, +infinity)

Find the fundamental period of sin(5t+[tex]\pi[/tex])

Homework Equations





The Attempt at a Solution



f(x)=sin(5t+[tex]\pi[/tex])

f(x)=f(x+2p)

f(x+2p)=sin(5t+[tex]\pi[/tex]+2p)

f(x+2[tex]\pi[/tex]/5)=sin(5t+[tex]\pi[/tex]+2[tex]\pi[/tex]/5)

The period i got was 2[tex]\pi[/tex]/5

f(x+2[tex]\pi[/tex]/5)=sin(5t+7[tex]\pi[/tex]/5)

f(x+2[tex]\pi[/tex]/5)=sin5tcos7[tex]\pi[/tex]/5 + cos5tsin7[tex]\pi[/tex]/5

What should I do next?

I dont know how can I go back to the original function.
 

Answers and Replies

  • #2
33,719
5,417
The period of y = sin(t) is 2pi, right?
The graph of y = sin(At) is a compression of y = sin(t) by a factor of A toward the y axis if A > 1, and is the expansion away from the y axis by a factor of 1/A if A < 1.

So for example, y = sin(3t) is compressed toward the y axis by a factor of 3, with the result that its period is 2pi/3. Your function can be written as f(x) = sin(5(t + pi/5)). All that the pi/5 represents is a translation to the left of the graph of y = sin(5t) by pi/5 units, so doesn't affect the period at all.

Nuff said?
 
  • #3
164
0
where does pi/5 came from? what do I do to the 2pi/5 period?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Sin(x) has period [itex]2\pi[/itex]. That means one period starts when x= 0 and ends when [itex]x= 2\pi[/itex]. For [itex]sin(5t+\pi)[/itex], then, one period starts when [itex]5t+ \pi= 0[/itex] and ends when [itex]5t+\pi= 2\pi[/itex]. From [itex]5t+\pi= 0[/itex], [itex]5t= -\pi[/itex] so [itex]t= -\pi/5[/itex] and from [itex]5t+\pi= 2\pi[/itex], [itex]5t= \pi[/itex] so [itex]t= \pi/5[/itex]. That means that one period of [itex]sin(5t+ \pi)[/itex] starts at [itex]-\pi/5[/itex] and ends at [itex]\pi/5[/itex], for a total length of [itex]\pi/5-(-\pi/5)= 2\pi/5[/itex]. Yes, the period is [itex]2\pi/5[/itex], just as you say.

Now, your main question appears to be showing that, in fact, [itex]f(t+ 2\pi/5)= f(t)[/itex]. Your error is that you just added [itex]2\pi/5[/itex] to the argument of sin rather than to x.

That is, you do NOT want "[itex]sin(5t+\pi+ 2\pi/5)[/itex]" .

You want, rather, [itex]sin(5(x+ 2\pi/5)+ \pi)[/itex][itex]= sin((5x+ 2\pi)+ \pi)[/itex][itex]= sin((5x+ \pi)+ 2\pi)[/itex]. Using the sum formulas as you do now gives [itex]sin(5(x+2\pi/5)+\pi)[/itex][itex]= sin((5x+\pi)cos(2\pi)+ cos(5x+\pi)sin(2\pi)[/itex]. And, because, of course, [itex]cos(2\pi)= 1[/itex] and [itex]sin(2\pi)= 0[/itex], that is just [itex]sin(5x+ \pi)[/itex] again.
 
Last edited by a moderator:
  • #5
164
0
so 2p will only be added to t since the part 5t determines the period? ok i understand what you said thanks...
 
  • #6
164
0
oh i see... so you regroup the 5t+2pi+pi into (5t+pi)+2pi?
how about if you dont regroup it?
 
  • #7
164
0
how about if the function comes in many terms?
for example: 1+cost+cos2t
Will I get the larger period among them or use the LCM?
if LCM, how do I do that?
 
  • #8
164
0
what should I do?
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,833
956
how about if the function comes in many terms?
for example: 1+cost+cos2t
Will I get the larger period among them or use the LCM?
if LCM, how do I do that?
In order that the function repeat, all parts must repeat which means you need the Least Common Multiple. But in this case, since the periods are [itex]\pi[/itex] and [itex]2\pi[/itex], that is the same as the largest period!

If the periods were, say, [itex]\pi/2[/itex] and [itex]3\pi/4[/itex], then the period of the sum would be [itex]3\pi/2[/itex].
 
  • #10
164
0
how do I can solve for the LCM if the given is fraction?
 
  • #11
33,719
5,417
how do I can solve for the LCM if the given is fraction?
See HallsOfIvy's last post where several periods are given that are fractional parts of pi. If that's not what you're asking, you need to be more specific.
 
  • #12
164
0
Im talking about his last post, pi/2 and 3pi/2, how it became 3pi/2. All I know about LCM is it for whole number just like the LCM of 3,5,and 15 is 30. For fraction, it is LCD.
 
  • #13
33,719
5,417
The LCM of pi/2 and 3pi/2 is 3pi/2, for the same reason that the LCM of 3 and 9 is 9. For the LCM of a group of numbers, you want the smallest number that all the numbers evenly divide.
 
  • #14
164
0
a mistake. it should be pi/2 and 3pi/4 is 3pi/2 there LCM.
 
  • #15
33,719
5,417
What's the smallest number that pi/2 and 3pi/4 both divide evenly?

On a different matter, the use of punctuation would help you communicate more effectively.
darkmagic said:
it should be pi/2 and 3pi/4 is 3pi/2 there LCM.
This is two sentences masquerading as one, so is harder to understand than is necessary.
 
  • #16
164
0
Am i correct?
the multiples of pi/2 are pi/2, pi, 3pi/2, 2pi....
the multiples of 3pi/4 are 3pi/4, 3pi/2,9pi/4...
so the LCM is 3pi/2.
 
  • #18
164
0
without using the listing of multiples, will this work?
lets say, find the LCM of pi/2 and 3pi/4
Using LCD pi/2 becomes 2pi/4 and 3pi/4 is still 3pi/4
By multiplying 2 and 3, it becomes 6
Then the LCM is 6pi/4 or 3pi/2
 
  • #19
164
0
is my last post correct?
 
  • #20
33,719
5,417
Yes, but you could have verified this by listing the multiples of pi/2 and 3pi/4.
 

Related Threads on The Fundamental Period

  • Last Post
Replies
7
Views
9K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
2
Views
559
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Top