Find Max Potential Energy per Unit Length of Wire

AI Thread Summary
The discussion revolves around determining the position along a vibrating wire where the potential energy per unit length is maximized. The mathematical analysis suggests that critical points occur at the endpoints (x = 0 and x = L) and the midpoint (x = L/2), but there is confusion regarding the maximum potential energy location. The textbook claims the maximum occurs at the midpoint, while the calculations indicate otherwise. Participants emphasize the importance of visualizing the physical behavior of the vibrating wire, noting that potential energy is highest when the displacement is maximum. Understanding the relationship between potential and kinetic energy in the context of the wire's motion is crucial for resolving the discrepancy.
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Homework Statement



A wire is fixed at both ends vibrating fundamentally. For what value of x (x position on the wire, with 0 being one edge and L being the other) is the potential energy per unit length has the maximum value?

Known: Length of wire (L), Tension in wire (T), Mass of wire (m), Amplitude (A).

Homework Equations



y(x,t) = A sin(kx) cos(wt)

dU/dx ~= T (dy/dx)^2 /2
(dy/dx is a partial derivative)

The Attempt at a Solution



dy/dx = Ak cos(kx) cos(wt)

dU/dx = T (Ak)^2 cos(wt)^2 cos(kx)^2

d/dx(dU/dx) = -T (Ak)^2 cos(wt)^2 2k cos(kx) sin(kx)
d/dx(dU/dx) = -T (Ak)^2 cos(wt)^2 k sin(2kx)

Critical points exist at d/dx(dU/dx) = 0, so since everything else is a constant, sin(2kx) = 0, and thus 2kx = 0 (x = 0), 2kx = pi (x = L/2), and 2kx = 2pi (x = L) are all solutions.

dU/dx = 0 at x = L/2, but when cos(wt)^2 > 0 then dU/dx is greater at x = 0, L than at x = L/2

The textbook says that the maximum potential energy per unit length occurs at the middle of the wire. My math says that it's at the two endpoints? Why?

Also, I googled the problem and they are taking dy/dx at 0 to get the result. I don't understand this. Can anyone explain?
 
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LoadedAnvils said:
The textbook says that the maximum potential energy per unit length occurs at the middle of the wire. My math says that it's at the two endpoints? Why?
What does the potential energy per unit length depend on? Where would these things be maximum?
Also, I googled the problem and they are taking dy/dx at 0 to get the result.
Who is "they"?
 


The potential energy per unit length depends on the partial derivative dy/dx (at least as an approximation) and the tension (which is constant).

Where would these things be maximum?

The partial dy/dx seems to be maximum at x = 0.

Who is "they"?

I looked here on page 78. This "solution" is what I don't understand. (I had worked out this problem without checking the solution to it first).

http://www.uccs.edu/~rtirado/Ch16%20ISM.pdf
 
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well ... I see you are stuck on thinking in terms of equations.
Have a look at a string that is vibrating like that and use physics.

Each length-element of the string is in constant motion, up and down.
So each mass element on the string is constantly exchanging kinetic and potential energy. When y is a maximum, the string between x and x+dx has maximum potential energy for that bit, when y=0, then the kinetic energy is a maximum. From this, without reference to any equations, what value of x would you expect to find the bit of the string that gets the most potential energy?

Note:
The reference in the link does not have a page 78 ... I found page 1678, but if any of the examples there used dy/dx I did not see it. There is a problem 78 too - but also not apropos. Some of the problems used dU/dx etc though but I stopped hunting.

The books pattern of getting you to "picture the problem" first is a good one for this problem too.
 
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