Transverse Wave: Time difference between two points

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SUMMARY

The discussion focuses on calculating the time difference between two arrivals of a transverse wave described by the function y(x,t) = 0.350sin(1.25x + 99.6t) at the point x=0 and height y=0.175m. The correct approach involves using the oscillation function y(0,t) = 0.350sin(99.6t) and finding the smallest two positive values of t that satisfy the equation. The wave speed is calculated as v = 79.68 m/s, leading to a distance covered of Δx = 1.68m during the time difference of 21 ms, as confirmed by the book's answer.

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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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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