Transverse Wave: Time difference between two points

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Homework Help Overview

The discussion revolves around a transverse wave described by the function y(x,t) = 0.350sin(1.25x + 99.6t) and focuses on determining the time difference between two instances when the wave reaches a specific height at a given point along the wire. The problem is situated within the context of wave mechanics and oscillatory motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the oscillation function for the specific point and height, questioning how to derive the time values corresponding to the wave reaching that height. There is discussion about using different values of k in the sine function to find multiple time instances and the reasoning behind selecting these values.

Discussion Status

Participants are actively engaging with the problem, raising questions about the methodology for finding time differences and the implications of using multiple formulas. There is a recognition that both sets of solutions need to be considered to find the smallest positive time values, but no consensus has been reached on the best approach.

Contextual Notes

Some participants express uncertainty about the theoretical background necessary to understand why certain methods are used, particularly regarding the selection of k values and the interpretation of the wave's oscillation behavior.

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Homework Statement



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?

Homework Equations



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

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.
 
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Darthkostis said:
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.
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.
 
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Darthkostis said:
of x = 0 at the height y =
I assume you meant at x=0 of the height y=...
 
haruspex said:
I assume you meant at x=0 of the height y=...

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.

haruspex said:
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.

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?
 
Darthkostis said:
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?
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.
 
haruspex said:
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.

Oh alright. So it's not something "set", I'm just looking for the smallest ones.

Thanks!
 

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