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Finding amplitude from superposition

  1. May 29, 2015 #1
    1. The problem statement, all variables and given/known data
    so there is this problem : find the amplitude from the superposition of Z1 and Z2 where
    Z1 : 8 sin 100t, A1= 8m
    and Z2 : 6 sin (100t-pi/2), A2=6m

    2. Relevant equations
    i know that all we need to do is add them and do the trig using sine additon, but i couldnt do that because the amplitude is different.
    3. The attempt at a solution
    yeah they give me the key answer but i dont understand because the solution only stated that all you need to do was do pythagorean theorem based on each amplitude.
     
  2. jcsd
  3. May 29, 2015 #2
    $$ \sin(100t-\pi/2) + \cos(100t) = ??? $$
     
  4. May 29, 2015 #3
    pardon me i dont understand where does the cos come from?
     
  5. May 29, 2015 #4
    Argumets are differs by ##\pi/2##, and functions are orthogonal, thus... Pythagoras.
     
  6. May 29, 2015 #5

    ehild

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    What do you get if you apply the addition law to sin(100t-pi/2)?

    Or even simpler: You know that sin(pi/2-x) = cos(x) And sin(-x)=sin(x)

    Edit: I think I was sleeping... Of course, sin(-x)=-sin(x) .
     
    Last edited: May 29, 2015
  7. May 29, 2015 #6
    sorry isnt sin -(x) = - sin x ?
     
  8. May 29, 2015 #7

    ehild

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    What do you mean with sin - (x)? Sine must have an argument. It is a function. You can not subtract something from "sin"
     
  9. May 29, 2015 #8
    sorry i mean sin(-x) = -sin(x)
     
  10. May 29, 2015 #9

    ehild

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    So sin(-x)=-sin(x). sin(x) is an even function. sin(pi/2)=1 and sin(-pi/2) = -1.
    Edit: Still sleeping, and forgetting English.
    Of course sine is odd.
     
    Last edited: May 29, 2015
  11. May 29, 2015 #10
    sorry can you submit your work here? i got stuck
     
  12. May 29, 2015 #11
    yeah thank you, so how do you solve this case?
    i change the second wave function (Z2) from 6sin(100t-pi/2) to -6cos(100t), so i have this problem :
    8sin(100t) - 6cos(100t), and i got stuck right there.
     
  13. May 29, 2015 #12
    Energy conservation: $$ A_{12}^2 = A_1^2 + A_2^2 $$ because there are orthogonal.
     
    Last edited: May 29, 2015
  14. May 29, 2015 #13

    ehild

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    You have to write 8 sin(100t)-6cos(100t)in the form A sin(100t + Φ ) where Φ is the phase constant. You can expand it, applying the addition rule for sine. What do you get?
    The equation has to be valid at every time, when 100t = 0 and when 100t=pi/2. What equations you get for Φ?
     
  15. May 29, 2015 #14
    This is the full solution that gives you two equation for ##A## and ##\phi##.
     
  16. May 29, 2015 #15

    BvU

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    Ehild ! sin(x) is an odd function !
    And Adi2 only asked in post 6 because of the typo in your post 5: the sin(-x)=sin(x) of course was meant to be ##\sin(-x)= -\sin x##

    Adi: I always like to make a picture. Better than many words and useful to check the outcome.

    Sines_addition.jpg
     
  17. May 29, 2015 #16

    ehild

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    Thank you BvU to pointing out my errors. I must have been very confused this morning...

    I wanted to make the OP recognize that A sin(Φ) = -6 and Asin(pi/2+Φ) = Acos(Φ) = 8, and he can get A2 by adding the squares.
     
    Last edited: May 29, 2015
  18. May 31, 2015 #17
    so i get this : A (sin100tcosΦ + sinΦcos100t)
    when 100t=0, i get this : A sin Φ
    when 100t=pi/w, i get this : A cos Φ
    so how do i find the Φ ? thank you for being patient because im veeery slow at trigs.
     
  19. May 31, 2015 #18

    ehild

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    Remember it was 8 sin(100t)-6cos(100t)=A sin(100t + Φ )
    What is the right side when 100t=0? What is it when 100t=pi/2? So what equations hold for cosΦ and sinΦ?
     
    Last edited: May 31, 2015
  20. May 31, 2015 #19

    BvU

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    Oh boy, here's me interfering again. So sorry!

    Adi has already shown what the right-hand side is, so now he/she should evaluate the left-hand side :smile:.

    But both Ehild and I mean the same thing: "I get ##A\sin\phi##" is incomplete: you were intended to get ## ... = A\sin\phi## with the emphasis on the = sign.
    The ##A\sin\phi## is a value, but ## ... = A\sin\phi## is an equation !
     
  21. May 31, 2015 #20

    ehild

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    Yes, I meant equations when 100t = 0 and when 100t = pi/2. What equations you get for Φ? (I asked in Posts #13 and #18). You got the right sides, but they must be equal to the left sides.
     
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