anonymous
  • anonymous
find the period and the amplitude of the function y=6cospix
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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JamesJ
  • 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?
eyust707
  • eyust707
in other words: cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 see a pattern?
JamesJ
  • 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.

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More answers

anonymous
  • anonymous
6?
eyust707
  • 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
JamesJ
  • JamesJ
6 is the amplitude, yes. What's the period.
JamesJ
  • JamesJ
@eyust707, you've got this one. Thanks.
eyust707
  • eyust707
any time James
anonymous
  • anonymous
i dont understand the period
anonymous
  • anonymous
cos(a x) any thing in front of x , in this case divide period= 2pi/a so cos(2 x) period = 2pi/2 = pi
JamesJ
  • 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)?
JamesJ
  • JamesJ
As eyust noted above, cos(0) = 1 cos(2pi) = 1 cos(4pi) = 1 cos(6pi) = 1 So what is T?
JamesJ
  • JamesJ
i.e., what is the period for the function g(x) = cos(x)?
JamesJ
  • 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) ?
JamesJ
  • 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?
anonymous
  • anonymous
ooh i think i understand now
anonymous
  • anonymous
lets say for y(x)=-2cos4x the period would be 2pi?
JamesJ
  • JamesJ
No.
JamesJ
  • 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.
JamesJ
  • 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.
JamesJ
  • 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
anonymous
  • anonymous
OMG this is hard!
JamesJ
  • JamesJ
No, it's just new for you. Do it a few more times and it'll be easy for you.
anonymous
  • anonymous
My teacher never taught me this and I'm trying to do the homework using the book but I find it really complicated
JamesJ
  • 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.
anonymous
  • anonymous
yes but that is very simple math lol
JamesJ
  • JamesJ
For me, the questions you're asking are also very simple.
JamesJ
  • JamesJ
but there was a time when they were new for me too. Just stick with it, and do a few more problems!
anonymous
  • anonymous
are you a math teacher?
JamesJ
  • JamesJ
Former University Lecturer
anonymous
  • anonymous
awesome
anonymous
  • anonymous
for y(x)=-2cos4x the amplitude is 2, correct?
JamesJ
  • JamesJ
Yes, the amplitude of y(x) = -2cos(4x) is 2, correct.
anonymous
  • anonymous
how do i find the frequency?
JamesJ
  • JamesJ
What's the definition of frequency, f?
anonymous
  • anonymous
the rate?
JamesJ
  • JamesJ
the rate of what?
anonymous
  • anonymous
of the wave
JamesJ
  • 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?
anonymous
  • anonymous
ooohh so the period of y(x)=-2cos4x is pi/2? because 2pi/4 is pi/2
JamesJ
  • JamesJ
yes.
anonymous
  • anonymous
omg yay lol
JamesJ
  • JamesJ
So now, returning to my last post and your question on frequency. What is the relationship between period T and frequency f?
anonymous
  • anonymous
1? because 1/2 times 1 is 2
anonymous
  • anonymous
i mean 1 over 1/2 is 2
JamesJ
  • JamesJ
Yes, so f = 1/T. Or T = 1/f
anonymous
  • anonymous
T =period?
JamesJ
  • 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.
JamesJ
  • JamesJ
Yes T = period.
JamesJ
  • JamesJ
The lower the frequency, the higher the period. The higher the frequency, the lower the period.
anonymous
  • anonymous
so if the period is pi/2, f=1/(pi/2)?
JamesJ
  • JamesJ
Yes. If T = pi/2, then f = 2/pi.
JamesJ
  • JamesJ
Which means every unit of time there are 2/pi complete oscillations.
anonymous
  • anonymous
got it :D
anonymous
  • anonymous
I have another problem :/ E(t)=110cos(120pit-pi/3)?
JamesJ
  • JamesJ
So you should know enough now to try and figure this out. First, what's the amplitude?
anonymous
  • anonymous
110
JamesJ
  • JamesJ
Yes, exactly. Now remember that the period of cos and sin is \( 2\pi \). I.e., \[ \cos(x + 2\pi) = \cos(x) \ \ \ \ \hbox{ and } \ \ \ \sin(x + 2\pi) = \sin(x) \]
JamesJ
  • 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?
anonymous
  • anonymous
60?
JamesJ
  • JamesJ
Noo...
anonymous
  • anonymous
ooh no no no no sorry
JamesJ
  • JamesJ
If 120πT=2π, then T = ...
anonymous
  • anonymous
2pi/120pi
JamesJ
  • JamesJ
Simplify
anonymous
  • anonymous
pi/60pi=1/60=0.016666667
JamesJ
  • JamesJ
T = 1/60. Don't write the decimal expansion unless you really, really have to. Always messy.
JamesJ
  • 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.
anonymous
  • anonymous
yes i actually got it! aah thank you soooo much!
JamesJ
  • JamesJ
What is the frequency of E(t). What is the value of f for that function?
anonymous
  • anonymous
60
JamesJ
  • 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.
JamesJ
  • JamesJ
If the units of t are seconds, E(t) has a complete oscillation every 1/60 seconds.
anonymous
  • anonymous
alright
JamesJ
  • JamesJ
Try and answer the first part of this problem: http://openstudy.com/#/updates/4ed01d1de4b04e045af4e631
anonymous
  • anonymous
i think its 8 but im not sure

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