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- YumYum247

w

- jamiebookeater

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- YumYum247

@aaronq can i please take just a few mintues of your time?!?!?!? Pleaase check my work.

- YumYum247

This is how i did it.......
Given: 30Hz, t = 0.20sec, d = 12m
To find the velocity, i need to find the wavelength first. Since V = fXLambda
V = Displacement in distance/ Displacement in time
V = 12m/0.20sec
V = 60sec
Now that we have our Velocity of the wave, we can now substitute both the given and reseultant values into the Formula V= F X Lambda
Lambda = V/F
Lambda = 60m/sec/30Hz =
Lambda = 2
Therefore the wavelength is 2meters long.
To find the Number of Loops, we need to take the inverse of Frequency and plug-in what's already given to us.....so
F = N/Change in time
30 Hz (given)= N/0.20sec
0.20sec X 30 = N
6 = N
Therefore the number of loops are 6.

- Astrophysics

What is takes sec, is that suppose to be .20 sec?

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## More answers

- YumYum247

already given to us...0.20sec

- YumYum247

i've lablled all the given values at the top.

- Astrophysics

Ok I see, well in your question it didn't mention that so just making sure.

- YumYum247

^_^

- Astrophysics

So you need to do \[v = f \lambda = (30 Hz)(12m) = 360 m/s\], good I see you have \[v = \frac{ \Delta d }{ \Delta t } \implies \Delta d = v \Delta t\] which is, \[\Delta d = (360m/s)(0.20s) = 72m\] so far you agree?

- YumYum247

how did you come up with "12"

- YumYum247

"is used to make waves in a rope 12m long", i though the length of the rope is 12m long -__-

- Astrophysics

Actually you maybe right

- Astrophysics

I guess I was reading it wrong and thought the wavelength is 12m but I guess the rope is!

- YumYum247

ok so once i have the velocity and the displacement, do i foolow the same procedure to find the wavelength.......λ = F/V

- Astrophysics

So, we should use \[v = \frac{ \Delta d }{ \Delta t } = \frac{ 12m }{ 0.20s } = 60m/s\] this seems better, I think the question is just worded weirdly so it caused the confusion.

- Astrophysics

\[\lambda = \frac{ v }{ f } = \frac{ 60m/s }{ 30 Hz } = 2m\] that makes more sense

- YumYum247

that's what i was wondering.........360m/sec.....that's like sound breaking speed!!! :")

- Astrophysics

Yeah it was a bit confusing when I was writing it out

- YumYum247

yep that's what i got.....λ = 2m

- Astrophysics

Now you can just divide that by the length of the string, so we have \[\frac{ 12m }{ 2m } = 6\] that is our wavelength in the string

- Astrophysics

amount of wavelengths*

- YumYum247

yep....that's right...i got the same.....but hey do you know the trick of solving these kinds of peoblems?!?!?!?!

- Astrophysics

There's no real trick, just this problem is worded oddly as you noticed haha

- Astrophysics

Lots of practice I suppose

- YumYum247

no i'm talking in general....tho, cuz i see some of the next problems i have in the book seem kinds hard.....

- YumYum247

is there any difference between solving for fixed end and free end?!?!??!?

- Astrophysics

There's no trick to it, just takes practice, read the question carefully, put all your knowns/ unkowns down and see what you can use :)

- Astrophysics

Well I would have to see the question to see what exactly you're asking

- YumYum247

ok....Thank you sooooooooooooooomuch, Luv you to death....!!! Please stay cute!!! :")

- Astrophysics

Haha np :) have fun!

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