Question

A sine function can be used to model light waves green light has a wave length or period of about 520 nanometers(nm) which equation best models green light? help, please.

Answer 1

@IrishBoy123 sorry to bother you, but can you please help?

Answer 2

"wave length or period"
you are using nms but which one do you mean?

Answer 3

That just how the question is phrased, it's above me.

Answer 4

Okay, I got the answer. Do you think you could help me with another? sorry for the quick change.

Answer 5

look up the frequency of green light and work out what the question is asking!
from wiki
Color Wavelength Frequency Photon energy
green 495–570 nm 526–606 THz 2.17–2.50 eV

so it is wavelength, using an average for the green range

a typical model would look like \[y = \sin (\omega t) \ \ where \ \omega = 2 \pi f = \frac{2 \pi c}{ \lambda} = c \frac{2 \pi }{ 520} = c \frac{\pi }{ 260}\]

which **looks like** D but its just drivel if you ask me

do you recognise any of what i have written?

Answer 6

Yes, that is what I got!

Answer 7

Would you kindly help me with my last one,please?

Answer 8

go ahead

Answer 9

he water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs at 3:30 p.m. What is a cosine function that models the variation in inches above and below the water level as a function of time in hours since 9:15 a.m.?

Answer 10

let's draw something, stat with a basic cosine function and we canhack it from there

Answer 11

Isn't the answer 20*cos(x-3.5)+32, but I just need to explain it because I just found the answer in a study guide, but not the how to.

Answer 12

By the way thanks for helping, I just need to get this done by Thursday.

Answer 13

OK, plug x = 0, into that equation and see what you get
you should get 12, right? is that what you get?


i can explain the 20 and the 32 , off the bat. the waves go from 12 to 52, so average is 12 + (52-12)/2 = 32. so amplitude = 52 - 32 = 20. you should draw that and fit it to a cosine curve.

but the rest (bits inside the cosine function) will take a bit more time. if you are in a rush, it will not work.

Answer 14

Well thanks for helping with what you could. It;s kind of late in where I live (Texas). I should get some sleep soon.

Answer 15

i'll leave a method here, but this is not as easy as it might look

you start with the basic idea that the Tide T can be modelled as
\[T(x) = A \cos (\omega x + \phi) + B\]
there are 4 constants to fit. we can see that A, the amplitude is 20 and B, the shift is 32. that was shown above.

x is the number of hours since 9.15, as per the question. ø is the phase shift, you have one in your solution that looks incorrect to me.

so starting with \[y = 20 \cos (\omega x + \phi) + 32, \ \\ at \ x = 0 \\ 12 = 20 \cos (\omega * 0 + \phi) + 32\\ \phi = \pi\]
then at half way between 9.15 and 3.30, i will use x = 3 (THIS is something that you would be more precise about in deriving your equation, you should use x = 3 1/8, i am simplifying as this is just an example and it should make for neat multiples of pi)
\[32 = 20 \cos (3 \omega + \pi) + 32 \\cos (3 \ \omega + \pi) = 0\\ \omega = -\frac{\pi}{6}\]
then checking this answer at 3.30 when x = 6
\[T(6) = 20 \cos (6 *\frac{-\pi}{6} + \pi) +32 = 52 \]

if you apply this method but don't use the simplifying x = 3 and x = 6, use the exact numbers x = 3 1/8 and x = 6 1/4, you will get a very similar solution. dirtier as the pi multiples won't be exact.