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

- anonymous

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

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- chestercat

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- 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

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

- 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

6?

- 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

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

- JamesJ

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

- eyust707

any time James

- anonymous

i dont understand the period

- 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

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

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

- JamesJ

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

- 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

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

ooh i think i understand now

- anonymous

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

- JamesJ

No.

- 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

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

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

OMG this is hard!

- JamesJ

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

- 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

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

yes but that is very simple math lol

- JamesJ

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

- JamesJ

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

- anonymous

are you a math teacher?

- JamesJ

Former University Lecturer

- anonymous

awesome

- anonymous

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

- JamesJ

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

- anonymous

how do i find the frequency?

- JamesJ

What's the definition of frequency, f?

- anonymous

the rate?

- JamesJ

the rate of what?

- anonymous

of the wave

- 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

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

- JamesJ

yes.

- anonymous

omg yay lol

- JamesJ

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

- anonymous

1? because 1/2 times 1 is 2

- anonymous

i mean 1 over 1/2 is 2

- JamesJ

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

- anonymous

T =period?

- 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

Yes T = period.

- JamesJ

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

- anonymous

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

- JamesJ

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

- JamesJ

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

- anonymous

got it :D

- anonymous

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

- JamesJ

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

- anonymous

110

- 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

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

60?

- JamesJ

Noo...

- anonymous

ooh no no no no sorry

- JamesJ

If 120πT=2π, then T = ...

- anonymous

2pi/120pi

- JamesJ

Simplify

- anonymous

pi/60pi=1/60=0.016666667

- JamesJ

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

- 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

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

- JamesJ

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

- anonymous

60

- 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

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

- anonymous

alright

- JamesJ

Try and answer the first part of this problem:
http://openstudy.com/#/updates/4ed01d1de4b04e045af4e631

- anonymous

i think its 8 but im not sure

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