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The Fundamental Period

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Nov 14, 2009 #2

    Mark44

    Staff: Mentor

    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?
     
  4. Nov 14, 2009 #3
    where does pi/5 came from? what do I do to the 2pi/5 period?
     
  5. Nov 14, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: Nov 14, 2009
  6. Nov 14, 2009 #5
    so 2p will only be added to t since the part 5t determines the period? ok i understand what you said thanks...
     
  7. Nov 14, 2009 #6
    oh i see... so you regroup the 5t+2pi+pi into (5t+pi)+2pi?
    how about if you dont regroup it?
     
  8. Nov 14, 2009 #7
    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?
     
  9. Nov 14, 2009 #8
    what should I do?
     
  10. Nov 14, 2009 #9

    HallsofIvy

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    Staff Emeritus
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    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].
     
  11. Nov 14, 2009 #10
    how do I can solve for the LCM if the given is fraction?
     
  12. Nov 14, 2009 #11

    Mark44

    Staff: Mentor

    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.
     
  13. Nov 14, 2009 #12
    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.
     
  14. Nov 14, 2009 #13

    Mark44

    Staff: Mentor

    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.
     
  15. Nov 14, 2009 #14
    a mistake. it should be pi/2 and 3pi/4 is 3pi/2 there LCM.
     
  16. Nov 14, 2009 #15

    Mark44

    Staff: Mentor

    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.
    This is two sentences masquerading as one, so is harder to understand than is necessary.
     
  17. Nov 14, 2009 #16
    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. Nov 15, 2009 #17

    Mark44

    Staff: Mentor

    Yes, that's correct.
     
  19. Nov 15, 2009 #18
    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
     
  20. Nov 15, 2009 #19
    is my last post correct?
     
  21. Nov 15, 2009 #20

    Mark44

    Staff: Mentor

    Yes, but you could have verified this by listing the multiples of pi/2 and 3pi/4.
     
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