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eli123 Group Title

find the period and the amplitude of the function y=6cospix

  • 2 years ago
  • 2 years ago

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  1. JamesJ Group Title
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    You can figure out the period from one time cos(pi.x) = 1 to the next time. So cos(pi.x) = 1 when x = 0, as cos(pi.0) = cos(0) = 1. What is the next value of x for which cos(pi.x) = 1?

    • 2 years ago
  2. eyust707 Group Title
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    in other words: cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 see a pattern?

    • 2 years ago
  3. JamesJ Group Title
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    As for the amplitude, that is the maximum value of this function. y = f(x) = 6 cos(pi.x). What is it's maximum value? Hint: it occurs when cos(pi.x) has its maximum value.

    • 2 years ago
  4. eli123 Group Title
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    6?

    • 2 years ago
  5. eyust707 Group Title
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    Yep so basically like james said, the amplitude is = to the maxium valuse that we can make 6cos(pi*x) Since the 6 is a constant the only thing we can change is the x. We need to change the x to something that will make "cos(pi*x)" as big as possible. if we plug in all the possible values around the unit circle you will find that cos never gets bigger than 1

    • 2 years ago
  6. JamesJ Group Title
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    6 is the amplitude, yes. What's the period.

    • 2 years ago
  7. JamesJ Group Title
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    @eyust707, you've got this one. Thanks.

    • 2 years ago
  8. eyust707 Group Title
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    any time James

    • 2 years ago
  9. eli123 Group Title
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    i dont understand the period

    • 2 years ago
  10. imranmeah91 Group Title
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    cos(a x) any thing in front of x , in this case divide period= 2pi/a so cos(2 x) period = 2pi/2 = pi

    • 2 years ago
  11. JamesJ Group Title
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    The period of a function f is the smallest number T for which f(x + T) = f(x). For the function f(x) = 6 cos(pi.x), it is therefore the smallest number T such that 6cos(pi(x+T)) = 6cos(pi.x) that is cos(pi.(x+T)) = cos(pi.x) Now if the pi wasn't there, draw the function g(x) = cos(x). For what value of T is it the case that g(x + T) = g(x)? i.e., cos(x + T) = cos(x)?

    • 2 years ago
  12. JamesJ Group Title
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    As eyust noted above, cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 So what is T?

    • 2 years ago
  13. JamesJ Group Title
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    i.e., what is the period for the function g(x) = cos(x)?

    • 2 years ago
  14. JamesJ Group Title
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    It's clear that the period of g(x) = cos(x) is T = 2pi. Now that being the case, what is the period of the function f(x) where f(x) = 6 cos(pi.x) ? I.e., for what value of T is it the case that f(x+T) = f(x) cos(pi(x+T)) = cos(pi.x) ?

    • 2 years ago
  15. JamesJ Group Title
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    For example, for x = 0, cos(pi(0+T) = cos(pi.0) i.e., cos(piT) = cos(0) = 1 i.e., cos(pi.T) = 1. What is the smallest such number T so that is the case?

    • 2 years ago
  16. eli123 Group Title
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    ooh i think i understand now

    • 2 years ago
  17. eli123 Group Title
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    lets say for y(x)=-2cos4x the period would be 2pi?

    • 2 years ago
  18. JamesJ Group Title
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    No.

    • 2 years ago
  19. JamesJ Group Title
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    By definition, it is the smallest number T such that y(x+T) = y(x) i.e., -2 cos(4(x+T)) = - 2 cos(4x) i.e., cos(4x + 4T) = cos(4x) Now cos has period 2pi. Hence 4T = 2pi or T = pi/2. Therefore the period of y(x) is T = pi/2.

    • 2 years ago
  20. JamesJ Group Title
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    or in other words, as imran was just writing, if you have a function f1(x) = sin(ax) or f2(x) = cos(ax), as both sin and cos have period 2pi, it follows that the period of both f1 and f2 is 2pi/a.

    • 2 years ago
  21. JamesJ Group Title
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    For example, the period T of f1 is the smallest number T such that f1(x + T) = f1(x) i.e., sin(a(x+T)) = sin(ax) i.e., sin(ax + aT) = sin(ax) i.e., aT = 2pi, because the period of sin is 2pi i.e., T = 2pi/a

    • 2 years ago
  22. eli123 Group Title
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    OMG this is hard!

    • 2 years ago
  23. JamesJ Group Title
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    No, it's just new for you. Do it a few more times and it'll be easy for you.

    • 2 years ago
  24. eli123 Group Title
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    My teacher never taught me this and I'm trying to do the homework using the book but I find it really complicated

    • 2 years ago
  25. JamesJ Group Title
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    It's like the first time you saw algebra. It seemed hard, but now you can solve equations like 2x + 4 = 6 In your sleep.

    • 2 years ago
  26. eli123 Group Title
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    yes but that is very simple math lol

    • 2 years ago
  27. JamesJ Group Title
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    For me, the questions you're asking are also very simple.

    • 2 years ago
  28. JamesJ Group Title
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    but there was a time when they were new for me too. Just stick with it, and do a few more problems!

    • 2 years ago
  29. eli123 Group Title
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    are you a math teacher?

    • 2 years ago
  30. JamesJ Group Title
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    Former University Lecturer

    • 2 years ago
  31. eli123 Group Title
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    awesome

    • 2 years ago
  32. eli123 Group Title
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    for y(x)=-2cos4x the amplitude is 2, correct?

    • 2 years ago
  33. JamesJ Group Title
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    Yes, the amplitude of y(x) = -2cos(4x) is 2, correct.

    • 2 years ago
  34. eli123 Group Title
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    how do i find the frequency?

    • 2 years ago
  35. JamesJ Group Title
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    What's the definition of frequency, f?

    • 2 years ago
  36. eli123 Group Title
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    the rate?

    • 2 years ago
  37. JamesJ Group Title
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    the rate of what?

    • 2 years ago
  38. eli123 Group Title
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    of the wave

    • 2 years ago
  39. JamesJ Group Title
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    I'll tell you. If a function is periodic, i.e., oscillates, it has a period, T such that f(t+T) = f(t) The frequency is the number of complete oscillations per unit of time. For example if T = 1, then there would be one oscillation per unit of time, seconds say. I.e., f = 1. If T = 2, there would be 1/2 an oscillation per second. I.e., f = 1/2. If T = 1/2, there would 2 oscillations per second, f = 2. Given all that, what is the relation between T and f?

    • 2 years ago
  40. eli123 Group Title
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    ooohh so the period of y(x)=-2cos4x is pi/2? because 2pi/4 is pi/2

    • 2 years ago
  41. JamesJ Group Title
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    yes.

    • 2 years ago
  42. eli123 Group Title
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    omg yay lol

    • 2 years ago
  43. JamesJ Group Title
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    So now, returning to my last post and your question on frequency. What is the relationship between period T and frequency f?

    • 2 years ago
  44. eli123 Group Title
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    1? because 1/2 times 1 is 2

    • 2 years ago
  45. eli123 Group Title
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    i mean 1 over 1/2 is 2

    • 2 years ago
  46. JamesJ Group Title
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    Yes, so f = 1/T. Or T = 1/f

    • 2 years ago
  47. eli123 Group Title
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    T =period?

    • 2 years ago
  48. JamesJ Group Title
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    Hence the higher the period, the lower the frequency. Makes sense because if the period is longer, there can be less complete oscillations in a second. Or the lower the period, the higher the frequency.

    • 2 years ago
  49. JamesJ Group Title
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    Yes T = period.

    • 2 years ago
  50. JamesJ Group Title
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    The lower the frequency, the higher the period. The higher the frequency, the lower the period.

    • 2 years ago
  51. eli123 Group Title
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    so if the period is pi/2, f=1/(pi/2)?

    • 2 years ago
  52. JamesJ Group Title
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    Yes. If T = pi/2, then f = 2/pi.

    • 2 years ago
  53. JamesJ Group Title
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    Which means every unit of time there are 2/pi complete oscillations.

    • 2 years ago
  54. eli123 Group Title
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    got it :D

    • 2 years ago
  55. eli123 Group Title
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    I have another problem :/ E(t)=110cos(120pit-pi/3)?

    • 2 years ago
  56. JamesJ Group Title
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    So you should know enough now to try and figure this out. First, what's the amplitude?

    • 2 years ago
  57. eli123 Group Title
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    110

    • 2 years ago
  58. JamesJ Group Title
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    Yes, exactly. Now remember that the period of cos and sin is \( 2\pi \). I.e., \[ \cos(x + 2\pi) = \cos(x) \ \ \ \ \hbox{ and } \ \ \ \sin(x + 2\pi) = \sin(x) \]

    • 2 years ago
  59. JamesJ Group Title
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    So go back to first principles to find the period T of your new function E(t). It is the number T such that E(t+T) = E(t) i.e., \[ 110 \cos(120\pi(t+T)-\pi/3) = 110 \cos(120\pi t-\pi/3) \] i.e., \[ \cos(120\pi(t+T)-\pi/3) = \cos(120\pi t - \pi/3) \] i.e., \[ \cos(120\pi t - \pi/3 + 120\pi T) = \cos(120\pi t - \pi/3) \] i.e., \[ 120\pi T = 2\pi\] because the period of cos is \( 2\pi \) Hence T = ... what?

    • 2 years ago
  60. eli123 Group Title
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    60?

    • 2 years ago
  61. JamesJ Group Title
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    Noo...

    • 2 years ago
  62. eli123 Group Title
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    ooh no no no no sorry

    • 2 years ago
  63. JamesJ Group Title
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    If 120πT=2π, then T = ...

    • 2 years ago
  64. eli123 Group Title
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    2pi/120pi

    • 2 years ago
  65. JamesJ Group Title
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    Simplify

    • 2 years ago
  66. eli123 Group Title
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    pi/60pi=1/60=0.016666667

    • 2 years ago
  67. JamesJ Group Title
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    T = 1/60. Don't write the decimal expansion unless you really, really have to. Always messy.

    • 2 years ago
  68. JamesJ Group Title
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    Now that looked complicated, but you can always just read it off from the coefficient of t. Given E(t)=110cos(120pi.t-pi/3), the coefficient of t is 120pi. Hence the period is T = 2pi/120pi = 1/60.

    • 2 years ago
  69. eli123 Group Title
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    yes i actually got it! aah thank you soooo much!

    • 2 years ago
  70. JamesJ Group Title
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    What is the frequency of E(t). What is the value of f for that function?

    • 2 years ago
  71. eli123 Group Title
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    60

    • 2 years ago
  72. JamesJ Group Title
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    Yes, exactly. If the units of t are seconds, E(t) has 60 complete oscillations per second. If the units of t are days, then E(t) has 60 complete oscillations per day. Etc.

    • 2 years ago
  73. JamesJ Group Title
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    If the units of t are seconds, E(t) has a complete oscillation every 1/60 seconds.

    • 2 years ago
  74. eli123 Group Title
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    alright

    • 2 years ago
  75. JamesJ Group Title
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    Try and answer the first part of this problem: http://openstudy.com/#/updates/4ed01d1de4b04e045af4e631

    • 2 years ago
  76. eli123 Group Title
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    i think its 8 but im not sure

    • 2 years ago
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