Question

How can I find the amplitude of this wave?

Answer 1

What kind of wave?

Answer 2

This one.

Answer 3

Answer choices?

Answer 4

I don't understand your meaning.

Answer 5

What is your choice of answer

Answer 6

My choice of answer? The problem I am facing is that the amplitude is not the same across the different periods. I need to find a way to calculate the amplitude for the entire wave.

Answer 7

Did the teacher give a hint how to find amplitude. What math level is this?

There isn't a single fixed amplitude across the wave. You could take the average of the amplitudes, but you will lose information doing this.

Alternatively you can model this wave as a sum of sines and cosines using fourier theory : Any periodic function f(t) can be represented by an infinite sum of sine and/or cosine functions that are harmonically related.

Answer 8

How can you model the wave?

Answer 9

checking my book :)

Answer 10

dude this is some huuuuuuuuuuuge level maths xD

Answer 11

what is plotted against x-axis?time?

Answer 12

Yes.

Answer 13

I think the average is
\[\frac{1}{T}\int\limits_{0}^{T}f(t)dt\]

Answer 14

$$
\large f(x) \approx a_o + \sum_{n=1}^Na_n \cos (\frac{n\pi x}{L} )+b_n\sin (\frac{n\pi x}{L}) \\
where\\
a_0 = \frac{1}{2L} \int_{-L}^{L}f(x)dx \\
a_n = \frac{1}{L} \int_{-L}^{L}f(x)\cos (\frac{n\pi x}{L} )dx \\
b_n= \frac{1}{L} \int_{-L}^{L}f(x)\sin (\frac{n\pi x}{L} )dx \\
$$

Answer 15

dont you think its a question of approximation??
try to draw a line that cut both the blue part and the white part equally and take the average of individual amplitudes...

Answer 16

L stands for length of one period

Answer 17

What is n?

Answer 18

n is a shorthand for n=1,2,3,... up to N

Answer 19

\[a _{0} = \frac{ 1 }{ 0.24 }\int\limits_{-0.24}^{0.24} f(x)dx\]

Answer 20

Should it look something like this?

Answer 21

For a0

Answer 22

right, but use 24, not .24 , assuming you want hours

Answer 23

Okay.

Answer 24

maybe its easier if we approach this as a sum of two waves, like the two waves i posted in that picture

Answer 25

\[a_{n} = \frac{ 1 }{ 12 } f(x) \cos \frac{ 2\pi x }{ 12 }\]

Answer 26

Is that looking good?

Answer 27

the problem is we don't know what f(x) is equal to.

Answer 28

but i this this problem can be solved by summing two sine functions, one that has a large period (long cycle) and one that has a short period (fast cycle)

Answer 29

How do we go about doing that?

Answer 30

Maybe the solution is simpler, how old are you ozhobbits?or particularly are you in school or college?

Answer 31

I am in the highest class in year 12.

Answer 32

There's probably a catch to the question then..

Answer 33

lets start with a -cos function

Answer 34

-cos(2pi/L * x)

Answer 35

Like that?

Answer 36

Lets start with this function
$$ \large y= -\cos(\frac{2\pi}{12} x )+1.4$$

Answer 37

I approximated the midpoint there

Answer 38

Okay. Should I calculate the midpoint?

Answer 39

yes if you can see it better

Answer 40

Okay. It is 1.25.

Answer 41

this function is getting close to it
y = -.5 cos ( 2pi /12 * x) + 1.25 ) - ( .5 cos (2pi/24x +1.4)

Answer 42

Should the midpoint be changed for the second one.

Answer 43

that is a phase shift, i had to introduce that

Answer 44

what was the directions for the assignment?

Answer 45

im basically tweaking the function on a handheld graphing calculator

Answer 46

To discover the formula for the wave using harmonic constituents.

Answer 47

Okay. How did you calculate the phase shift?

Answer 48

2.05.

just grab the middle (median) peak for each of the bigger and smaller wave patterns, and then calc a simple average of these 2.

looks like Brisbane is going to be underwater soon, unless this pattern reverses.

Answer 49

\[\frac{ 1 }{ 24}\left[ x \right]_{-12}^{12} - \frac{ 1 }{ 12 }\left[ x \cos \frac{ 2pix }{ 12 } \right]_{-12}^{12} - \frac{ 1 }{ 12 }\left[ x \sin\frac{ 2pix }{ 12} \right]_{-12}^{12}\]