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The mechanical energy of an element of the string is :

$$dE(x,t) = \frac{1}{2}\left[T_0\left(\frac{\partial y}{\partial x}\right)^2 + \mu\left(\frac{\partial y}{\partial t}\right)^2\right]dx$$

Where $T_0$ is the tension at rest and $\mu$ is the linear mass density.

I am to find a differential equation between the linear energy density $e(x,t) = \frac{dE}{dx}$ and the power $P_{1\rightarrow 2} (x,t) $ received by the right part (2) of the string from the left part (1) .

How am I supposed to do that ?

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  • $\begingroup$ The equation you are looking for is integral-differential, actually... $\endgroup$
    – Valter Moretti
    Commented Mar 30, 2015 at 9:46
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    $\begingroup$ $\frac{d}{dt}\int_{x_1}^{x_2} e(t,x) dx = P_{x_2}(t) -P_{x_1}(t)$ $\endgroup$
    – Valter Moretti
    Commented Mar 30, 2015 at 9:49
  • $\begingroup$ Ok thanks, but how do you get there ? $\endgroup$
    – mwa1
    Commented Mar 30, 2015 at 10:22
  • $\begingroup$ It is not, Valter, due to locality. The energy moving through a point is a local quantity related to other local quantities by purely differential operations - so whenever there is an integral sign, it gets cancelled by a derivative. $\endgroup$
    – Luboš Motl
    Commented Mar 30, 2015 at 10:24
  • $\begingroup$ Yes Lubos, but as far as I understand the points 1 and 2 are the end points of the string, so an integration should take place. Perhaps I did not understand the question. $\endgroup$
    – Valter Moretti
    Commented Mar 30, 2015 at 11:13

1 Answer 1

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D'Alembert equation reads, for the considered case, $$\mu\frac{\partial^2 y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}\:.\tag{1}$$ It is nothing but $F=ma$ along the vertical direction ($y$). Here $y$ denotes the small deformation of the string along the vertical direction from the stationary (horizontal) configuration. We disregard horizontal deformations. $T_0 \frac{\partial^2 y}{\partial x^2}$ is the $x$-derivative of vertical component of force acting on a infinitesimal segment of string $dx$, when the deformation is small, assuming that the absolute value of the tension $T_0$ is constant. In other words $$f_y = T_0 \frac{\partial y}{\partial x}\tag{2}$$
is the $y$ component of either the tension or the force acting at the endpoints.

From (1) and the defintion of $e(t,x)$

$$e(t,x) = \frac{1}{2}\left[T_0\left(\frac{\partial y}{\partial x}\right)^2 + \mu\left(\frac{\partial y}{\partial t}\right)^2\right]$$ one easily sees (just computing derivatives) that $$\frac{\partial e}{\partial t} = \frac{\partial}{\partial x} \left(\frac{\partial y}{\partial t} T_0\frac{\partial y}{\partial x}\right)\:.\tag{3}$$ Integrating this identity along the string, from the point (maybe endpoint) $x_1$ to the point (maybe endpoint) endpoint $x_2$, we have $$\frac{d}{dt}\int_{x_1}^{x_2} e(t,x) dx = \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_2} - \left.\left(\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}\right)\right|_{x_1}\tag{4}$$ As $\frac{\partial y}{\partial t}$ is the vertical velocity of the endpoint, taking (2) into account, we conclude that $\frac{\partial y}{\partial t}T_0\frac{\partial y}{\partial x}$ is the power of the force acting on the endpoint (or at a generic point of the string).

Equations (3) and (4) establish relations, respectively local and global, between the density of energy and the power of the forces acting along the string.

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  • $\begingroup$ Actually I don't easily see how you got to equation (3). I'm stuck at $\frac{\partial e(x,t)}{\partial t} = T_0 \frac{\partial y}{\partial x} \frac{\partial \left(\frac{\partial y}{\partial x}\right)}{\partial t} + \mu \frac{\partial y}{\partial t}\frac{\partial ^2 y}{\partial t^2}$ $\endgroup$
    – mwa1
    Commented Mar 30, 2015 at 13:46
  • $\begingroup$ Use $\mu \frac{\partial^2y}{\partial t^2} = T_0 \frac{\partial^2 y}{\partial x^2}$ in the second term in the right-hand side and take advantage of $\frac{\partial \frac{\partial y}{\partial x}}{\partial t}= \frac{\partial^2 y}{\partial x \partial t}$ $\endgroup$
    – Valter Moretti
    Commented Mar 30, 2015 at 14:42
  • $\begingroup$ @Valter Moretti: Sir, I'll be grateful to you if you please explain to me how the wave-equation is analogous to the second law. And sir, 2nd derivative of position means acceleration; what do these 2nd derivatives mean? $\endgroup$
    – user36790
    Commented Apr 8, 2015 at 18:34

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