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Transverse Wave: Time difference between two points

  1. Jul 13, 2017 #1
    1. The problem statement, all variables and given/known data

    A transverse wave that is propagated through a wire, is described through this function: y(x,t) = 0.350sin(1.25x + 99.6t) SI

    Consider the point of the wire that is found at x= 0:

    a) What's the time difference between the two first arrivals of x = 0 at the height y = 0.175m?
    b) How much distance does the wave cover during that time?

    2. Relevant equations

    v = λ*f
    v = Δx/Δt
    sinx = sinφ => x = 2kπ + φ OR Χ = 2kπ + π - φ

    3. The attempt at a solution

    a) First up, the oscillation function for x = 0 is: y(0,t) = 0.350sin(99.6t)

    For y = 0.175m => ... 0,5 = sin(π/6) = sin(99,6t) => 99.6t = 2kπ + π/6 OR 99.6t = 2kπ + 5π/6

    And here's I find the problem. I don't remember how to solve these (it's been a while), so while I know that I should put k = 0, get a result, then k = 1, get a result, and then find the difference between the two, I don't know which formula to pick.

    For example, for k = 0 we have: t = 5,25 * 10-3s from the first, and t = 0,026s from the second.
    Likewise, for k = 1 we have: t = 0.068s & t = 0.089s

    The book's answer is t = 21 ms, which I get if I find the difference between the first set (0,026 - 0,00525 gives 0,7 ms), or the difference between the second set (0.089 - 0.068 gives us a perfect 0,021). Problem is, I don't know why. I don't remember the theory behind this is what I'm saying. Why can't I find the difference between the results of just one formula, one for k = 0, and the other for k = 1, eg 0,068 - 0,00525.

    b) That's an easy one. v = λ*f <=>... <=> v = 79.68 m/s | v = ΔX/ΔT <=> ... <=> Δx = 1.68m

    Any help is appreciated!

    PS: I then tried something different, and while I didn't get proper results, I'd like to know why it's wrong. Say, from putting x = 0 & y = 0.175 into the main function, if solved through arcsin, I can get t ~ 0,0052. So, considering that x goes from Position of Balance, to = +A, then to PoB, then to -A, then back to PoB and so on and so forth, couldn't I find the "wanted time" by this logic:

    Δt = (T/4 - t) + T/4 + T/4 + T/4 + t

    Obviously it's wrong, but I want to know why.
     
  2. jcsd
  3. Jul 13, 2017 #2

    haruspex

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    It is not a matter of picking one formula or the other. Both give instants at which y has the desired value. You want the smallest two positive values of t which satisfy either.
     
  4. Jul 13, 2017 #3

    haruspex

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    I assume you meant at x=0 of the height y=...
     
  5. Jul 13, 2017 #4
    Yeah, the "particle"/point that is found at x = 0 goes "up and down"/oscillates between +A/-A, and I need to find the time difference between the first time it arrives at y = 0.175 m and the second time it arrives there.

    Ah, so I have to take a constant, k in this instance, and pluck it in BOTH formulas? So, essentially, I need to use both formulas for just one k? How does that work though? Shouldn't I get the same result if I picked just one formula, and then just plucked in two different values for k (0 & 1)? What's the theory/backstory behind it?
     
  6. Jul 13, 2017 #5

    haruspex

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    The set of times at which y(0,t) will have the desired value is the union of the two sets of solutions you found. You want the smallest two positive values from that union. It could be the smallest two from set, the smallest two from the other set, or the smallest one from each.
     
  7. Jul 13, 2017 #6
    Oh alright. So it's not something "set", I'm just looking for the smallest ones.

    Thanks!
     
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