Question

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

Answer 1

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?

Answer 2

in other words:
cos(0) = 1
cos(2pi) = 1
cos(4pi) = 1
cos(6pi) = 1

see a pattern?

Answer 3

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.

Answer 4

6?

Answer 5

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

Answer 6

6 is the amplitude, yes.

What's the period.

Answer 7

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

Answer 8

any time James

Answer 9

i dont understand the period

Answer 10

cos(a x)

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

so cos(2 x)
period = 2pi/2 = pi

Answer 11

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)?

Answer 12

As eyust noted above,
cos(0) = 1
cos(2pi) = 1
cos(4pi) = 1
cos(6pi) = 1

So what is T?

Answer 13

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

Answer 14

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) ?

Answer 15

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?

Answer 16

ooh i think i understand now

Answer 17

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

Answer 18

No.

Answer 19

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.

Answer 20

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.

Answer 21

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

Answer 22

OMG this is hard!

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

yes but that is very simple math lol

Answer 27

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

Answer 28

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

Answer 29

are you a math teacher?

Answer 30

Former University Lecturer

Answer 31

awesome

Answer 32

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

Answer 33

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

Answer 34

how do i find the frequency?

Answer 35

What's the definition of frequency, f?

Answer 36

the rate?

Answer 37

the rate of what?

Answer 38

of the wave

Answer 39

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?

Answer 40

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

Answer 41

yes.

Answer 42

omg yay lol

Answer 43

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

Answer 44

1? because 1/2 times 1 is 2

Answer 45

i mean 1 over 1/2 is 2

Answer 46

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

Answer 47

T =period?

Answer 48

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.

Answer 49

Yes T = period.

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

got it :D

Answer 55

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

Answer 56

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

Answer 57

110

Answer 58

Yes, exactly.

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

Answer 59

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?

Answer 60

60?

Answer 61

Noo...

Answer 62

ooh no no no no sorry

Answer 63

If 120πT=2π, then T = ...

Answer 64

2pi/120pi

Answer 65

Simplify

Answer 66

pi/60pi=1/60=0.016666667

Answer 67

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

Answer 68

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.

Answer 69

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

Answer 70

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

Answer 71

60

Answer 72

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.

Answer 73

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

Answer 74

alright

Answer 75

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

Answer 76

i think its 8 but im not sure