## eli123 4 years ago find the period and the amplitude of the function y=6cospix

1. JamesJ

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. eyust707

in other words: cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 see a pattern?

3. JamesJ

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.

4. eli123

6?

5. eyust707

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

6. JamesJ

6 is the amplitude, yes. What's the period.

7. JamesJ

@eyust707, you've got this one. Thanks.

8. eyust707

any time James

9. eli123

i dont understand the period

10. imranmeah91

cos(a x) any thing in front of x , in this case divide period= 2pi/a so cos(2 x) period = 2pi/2 = pi

11. JamesJ

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)?

12. JamesJ

As eyust noted above, cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 So what is T?

13. JamesJ

i.e., what is the period for the function g(x) = cos(x)?

14. JamesJ

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) ?

15. JamesJ

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?

16. eli123

ooh i think i understand now

17. eli123

lets say for y(x)=-2cos4x the period would be 2pi?

18. JamesJ

No.

19. JamesJ

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.

20. JamesJ

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.

21. JamesJ

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

22. eli123

OMG this is hard!

23. JamesJ

No, it's just new for you. Do it a few more times and it'll be easy for you.

24. eli123

My teacher never taught me this and I'm trying to do the homework using the book but I find it really complicated

25. JamesJ

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.

26. eli123

yes but that is very simple math lol

27. JamesJ

For me, the questions you're asking are also very simple.

28. JamesJ

but there was a time when they were new for me too. Just stick with it, and do a few more problems!

29. eli123

are you a math teacher?

30. JamesJ

Former University Lecturer

31. eli123

awesome

32. eli123

for y(x)=-2cos4x the amplitude is 2, correct?

33. JamesJ

Yes, the amplitude of y(x) = -2cos(4x) is 2, correct.

34. eli123

how do i find the frequency?

35. JamesJ

What's the definition of frequency, f?

36. eli123

the rate?

37. JamesJ

the rate of what?

38. eli123

of the wave

39. JamesJ

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?

40. eli123

ooohh so the period of y(x)=-2cos4x is pi/2? because 2pi/4 is pi/2

41. JamesJ

yes.

42. eli123

omg yay lol

43. JamesJ

So now, returning to my last post and your question on frequency. What is the relationship between period T and frequency f?

44. eli123

1? because 1/2 times 1 is 2

45. eli123

i mean 1 over 1/2 is 2

46. JamesJ

Yes, so f = 1/T. Or T = 1/f

47. eli123

T =period?

48. JamesJ

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.

49. JamesJ

Yes T = period.

50. JamesJ

The lower the frequency, the higher the period. The higher the frequency, the lower the period.

51. eli123

so if the period is pi/2, f=1/(pi/2)?

52. JamesJ

Yes. If T = pi/2, then f = 2/pi.

53. JamesJ

Which means every unit of time there are 2/pi complete oscillations.

54. eli123

got it :D

55. eli123

I have another problem :/ E(t)=110cos(120pit-pi/3)?

56. JamesJ

So you should know enough now to try and figure this out. First, what's the amplitude?

57. eli123

110

58. JamesJ

Yes, exactly. Now remember that the period of cos and sin is $$2\pi$$. I.e., $\cos(x + 2\pi) = \cos(x) \ \ \ \ \hbox{ and } \ \ \ \sin(x + 2\pi) = \sin(x)$

59. JamesJ

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?

60. eli123

60?

61. JamesJ

Noo...

62. eli123

ooh no no no no sorry

63. JamesJ

If 120πT=2π, then T = ...

64. eli123

2pi/120pi

65. JamesJ

Simplify

66. eli123

pi/60pi=1/60=0.016666667

67. JamesJ

T = 1/60. Don't write the decimal expansion unless you really, really have to. Always messy.

68. JamesJ

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.

69. eli123

yes i actually got it! aah thank you soooo much!

70. JamesJ

What is the frequency of E(t). What is the value of f for that function?

71. eli123

60

72. JamesJ

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.

73. JamesJ

If the units of t are seconds, E(t) has a complete oscillation every 1/60 seconds.

74. eli123

alright

75. JamesJ