Find Max Potential Energy per Unit Length of Wire

[LoadedAnvils]

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?

 

[Simon Bridge]

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"?

 

[LoadedAnvils]

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

 

[Simon Bridge]

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.

 

[Nickie]

The textbook may be referring to the maximum potential energy per unit length at a specific point in time, rather than considering all points in time. Taking the derivative and setting it equal to 0 gives you the critical points, but it does not tell you whether those points correspond to a maximum, minimum, or saddle point. In this case, if you plug in the values for x = 0 and x = L/2, you will see that they both correspond to a minimum potential energy per unit length, while x = L corresponds to a maximum. This can be seen by plugging in the values for x into the equation for potential energy per unit length, dU/dx = T(Ak)^2 cos(wt)^2 cos(kx)^2. At x = 0 and x = L/2, the cosine terms are both equal to 1, while at x = L, the cosine term is equal to -1, resulting in a maximum value for potential energy per unit length.

Additionally, taking the derivative at x = 0 is not necessary for solving this problem. It is simply a shortcut that allows you to find the critical points more easily. However, as mentioned before, it does not tell you whether those points correspond to a maximum or minimum. To determine the maximum potential energy per unit length, you would need to plug in the values for x = 0, x = L/2, and x = L into the equation and compare the values.