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YumYum247
 one year ago
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YumYum247
 one year ago
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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1@aaronq can i please take just a few mintues of your time?!?!?!? Pleaase check my work.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1This 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 plugin 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
 one year ago
Best ResponseYou've already chosen the best response.1What is takes sec, is that suppose to be .20 sec?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1already given to us...0.20sec

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1i've lablled all the given values at the top.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok I see, well in your question it didn't mention that so just making sure.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.1how did you come up with "12"

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1"is used to make waves in a rope 12m long", i though the length of the rope is 12m long __

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Actually you maybe right

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I guess I was reading it wrong and thought the wavelength is 12m but I guess the rope is!

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1ok so once i have the velocity and the displacement, do i foolow the same procedure to find the wavelength.......λ = F/V

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So, 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\lambda = \frac{ v }{ f } = \frac{ 60m/s }{ 30 Hz } = 2m\] that makes more sense

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1that's what i was wondering.........360m/sec.....that's like sound breaking speed!!! :")

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah it was a bit confusing when I was writing it out

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1yep that's what i got.....λ = 2m

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Now 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
 one year ago
Best ResponseYou've already chosen the best response.1amount of wavelengths*

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1yep....that's right...i got the same.....but hey do you know the trick of solving these kinds of peoblems?!?!?!?!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1There's no real trick, just this problem is worded oddly as you noticed haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Lots of practice I suppose

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1no i'm talking in general....tho, cuz i see some of the next problems i have in the book seem kinds hard.....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1is there any difference between solving for fixed end and free end?!?!??!?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1There'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
 one year ago
Best ResponseYou've already chosen the best response.1Well I would have to see the question to see what exactly you're asking

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1ok....Thank you sooooooooooooooomuch, Luv you to death....!!! Please stay cute!!! :")

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Haha np :) have fun!
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