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 ?

  • $\begingroup$ The equation you are looking for is integral-differential, actually... $\endgroup$ Mar 30, 2015 at 9:46
  • 1
    $\begingroup$ $\frac{d}{dt}\int_{x_1}^{x_2} e(t,x) dx = P_{x_2}(t) -P_{x_1}(t)$ $\endgroup$ Mar 30, 2015 at 9:49
  • $\begingroup$ Ok thanks, but how do you get there ? $\endgroup$
    – mwa1
    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. $\endgroup$ 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. $\endgroup$ Mar 30, 2015 at 11:13

1 Answer 1


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.

  • $\begingroup$ 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
    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 $\frac{\partial \frac{\partial y}{\partial x}}{\partial t}= \frac{\partial^2 y}{\partial x \partial t}$ $\endgroup$ 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
    Apr 8, 2015 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.