## YumYum247 one year ago w

1. YumYum247

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

2. 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.

3. Astrophysics

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

4. YumYum247

5. YumYum247

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

6. Astrophysics

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

7. YumYum247

^_^

8. 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?

9. YumYum247

how did you come up with "12"

10. YumYum247

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

11. Astrophysics

Actually you maybe right

12. Astrophysics

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

13. YumYum247

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

14. 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.

15. Astrophysics

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

16. YumYum247

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

17. Astrophysics

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

18. YumYum247

yep that's what i got.....λ = 2m

19. 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

20. Astrophysics

amount of wavelengths*

21. YumYum247

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

22. Astrophysics

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

23. Astrophysics

Lots of practice I suppose

24. YumYum247

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

25. YumYum247

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

26. 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 :)

27. Astrophysics

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

28. YumYum247

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

29. Astrophysics

Haha np :) have fun!