Energy of a vibrating string (continuous field)by flix Tags: continuous, energy, field, string, vibrating 

#1
Sep1708, 09:35 AM

P: 13

Hello,
I know this has already been asked (unfortunately without answer)... learning once again for an exam (quantum field theory) I can't figure out a feature of a very central quantity: the total energy of a vibrating string. Let's start at the string (field) wave equation: [tex] \frac{1}{c^2} \, \frac{\partial^2 \Phi(x,t)}{\partial t^2} = \frac{\partial^2 \Phi(x,t)}{\partial x^2} [/tex] With proper boundary conditions you get a set of modes, for example [tex] \Phi_r(x,t) = A_r(t) \, sin(\frac{r \pi x}{L}) [/tex] and with that the general motion of the string: [tex] \Phi(x,t) = \sum\limits_{r=1}^\infty A_r(t) \, sin(\frac{r \pi x}{L}) [/tex] Now, my book confronts me with the total energy of the vibrating string, declaring it as "analogous to the discrete summation": [tex]E = \int\limits_0^L \left[ \frac{1}{2} \rho {(\frac{\partial \Phi}{\partial t})}^2 + \frac{1}{2} \rho c^2 {(\frac{\partial \Phi}{\partial x})}^2 \right] \, dx , \,\,where\, the \, "discrete"\, case\, was \,\,\,\,\,\,E = \sum\limits_{r=1}^N \frac{1}{2} m \dot{q}_r ^2 + V(q_1, ..., q_N) [/tex] I can't work this out. I understand that the first term is the kinetic energy, alright. But the second term I don't get. The potential Energy / Length on the string is [tex] E_{pot} / L =  \frac{1}{2 L} D \, \Phi ^2 =  \frac{1}{2} \omega ^2 \, \rho \, \Phi ^2 [/tex] as far as I can see from classical mechanics ([tex] D \, \Phi [/tex] is the force pulling the "mass element" backwards), but why the derivative [tex] {(\frac{\partial \Phi}{\partial x})} [/tex] insted of just [tex] \Phi [/tex] ? Can somebody help me out? I already wasted too much time on this... I hate it when I don't understand stuff from years back that I just should know. 



#2
Sep1708, 09:55 AM

Sci Advisor
P: 8,005

Because the displacement from "equilibrium" of one element is taken relative to the element just next to it. An element doesn't want to be at absolute zero, it just wants to be flat relative to its surrounding elements.
So for example, let's say you fix your string at both ends, then set it so that it rises, remains flat for a long length, then goes back to zero. The whole flat portion doesn't contribute to the energy, because there is no tension pulling an element there back down, so it doesn't "know" that it has been displaced. 



#3
Sep1708, 10:34 AM

P: 13

I see what you are heading at, still I don't get it mathematically.
The true Force pulling the string back is: [tex]F(x) = T \, \frac{\partial ^2 \Phi}{\partial x^2} [/tex] with [tex]T = \rho c^2 [/tex] being the tension (pulling at the ends of that string segment tangentially). So the potential energy of that string segment should be [tex] E(x) = \int\limits_0^{\Phi} \rho c^2 \frac{\partial ^2 \Phi}{\partial x^2} \, dx[/tex] and, unluckily, this is NOT [tex] E(x) = \frac{1}{2} \, \rho c^2 (\frac{\partial \Phi}{\partial x})^2 [/tex] since [tex] \frac{\partial}{\partial x} \, (\frac{\partial \Phi}{\partial x})^2 = 2 \frac{\partial \Phi}{\partial x} \, \frac{\partial ^2 \Phi}{\partial x^2} [/tex] Am I too stupid to differentiate today? I just can't find the error.... 



#4
Sep1708, 10:46 AM

P: 13

Energy of a vibrating string (continuous field)
damn, stupid mistake.
of course I don't have to integrate the force over x, but over [tex]\Phi[/tex] ! One gets: [tex]E(x) = \int\limits_0^{\Phi} \rho c^2 \frac{\partial^2 \Phi}{\partial x^2} \,d\Phi[/tex] and this looks much more like it might ultimatel lead to victory if I can somehow find my old differential calculus knowledge again :P 



#5
Sep1708, 11:06 AM

Sci Advisor
P: 8,005





#6
Sep1708, 11:23 PM

Sci Advisor
P: 8,005

I am still stuck. Physically it seems obvious that it should be:
dF~kdy dU~k(dy)^{2} But I can't get there without fudging. 



#7
Sep1808, 02:56 AM

P: 13

yeah I don't get there, either.
I'll have another go today and I will post any progress here. Damn you, classical mechanics! 



#8
Sep1808, 03:52 PM

Sci Advisor
P: 817

[QUOTE]
[tex]\mathcal{L} = \frac{1}{2} \{ (\frac{\partial \Phi}{\partial t})^{2}  (\frac{\partial \Phi}{\partial x})^{2}\}[/tex] Now, calculate the energy from [tex] E = \int \ dx \left( \frac{\partial \mathcal{L}}{\partial \dot{\Phi}} \dot{\Phi}  \mathcal{L} \right)[/tex] regards sam 



#9
Sep1808, 04:11 PM

P: 555

Possibly one way would be the decomposition of the shape of the string into its harmonics. Then you just take an average with weights given by the Fourier coeficients, of the energy of each harmonic present.
The energy of an harmonic can be taken ( I guess ) by taking the sinousoidal shape of the string when all its elements are passing by the x axis. In this situation all energy must be in kinetic form. You must simply integrate the kinetic of each element to find the total energy. I am convinced of this method. This may be dangerous... if someone has some refutation, please show here. Best wishes DaTario 



#10
Sep2008, 06:38 AM

P: 13

@samalkhaiat: yeah, well, this is exactly why I want to understand this, to motivate the Lagrangian ..
ok, for all other still interested, I found a classical derivation. Of course my initial assumption was plain wrong, as pointed out by atyy: the potential energy of a string element does NOT depend on its total displacement but only on the relative position to its "neighbour elements", since the real force runs along the string (string tension). This is why my integral idea fails. proper classical derivation: discrete string element displacement: [tex]u_i(x_i, t)[/tex] distance of 2 string elements, each of size [tex] \Delta x[/tex]: [tex]\Delta s = \sqrt{(\Delta x)^2 + (u_{i+1}  u_i)^2 } \approx \Delta x \left(1+ \frac{(u_{i+1}  u_i)^}{2 (\Delta x)^2} \right) [/tex] Discrete pot. energy for string "ground" tension T: [tex]U = \sum_{i=1}^N T \Delta x \frac{(u_{i+1}  u_i)^}{2 (\Delta x)^2} [/tex] In the limit [tex]N \longrightarrow \infty[/tex] : [tex]U = \lim_{N\to\infty} \sum_{i=1}^N T \Delta x \frac{(u_{i+1}  u_i)^}{2 (\Delta x)^2} = \frac{T}{2} \int\limits_0^L \left( \frac{\partial u(x,t)}{\partial x} \right) ^2 \,dx [/tex] 


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