I was solving some exercises on energy transportation on waves and found myself in trouble. The question was "Determine the average energy density in the wave, $\left\langle \frac{dE}{dx} \right\rangle$, in Joules per meter up to two decimal places". Earlier it had given the informations that we were dealing with a sinusoidal wave on a string, also giving the values of the wave velocity ($v = 20$ m/s), the linear density of the string in which the wave was propagating ($\mu = 0.20$ kg/m), the wave's amplitude ($A = 0.10$ m) and it's frequency ($\nu = 5.0$ Hz). After some calcuation, I found that the wavelenght is $\lambda = 4.0$ m and that the tension force applied to the string is $F_T = 80$ N. I solved the question like this:

$$\frac{dK}{dx} = \frac{1}{2} \mu \left( \frac{\partial y}{\partial t} \right)^2 \qquad \frac{dU}{dx} = \frac{1}{2} F_T \left( \frac{\partial y}{\partial x} \right)^2$$

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

$$\frac{dE}{dx} = \frac{1}{2} \mu \left( \frac{\partial y}{\partial x} \right)^2 \cdot \left( \frac{dx}{dt} \right)^2+ \frac{1}{2} F_T \left( \frac{\partial y}{\partial x} \right)^2$$

$$\frac{dE}{dx} = \frac{1}{2} \mu \left( \frac{\partial y}{\partial x} \right)^2 \cdot v^2+ \frac{1}{2} F_T \left( \frac{\partial y}{\partial x} \right)^2$$

$$F_T = \mu \cdot v^2$$

$$\frac{dE}{dx} = F_T \left( \frac{\partial y}{\partial x} \right)^2$$

$$y(x,t) = A \cos (\phi) \qquad \phi = kx - \omega t + \delta$$

$$\frac{\partial y}{\partial x} = - A k \sin (\phi)$$

$$\frac{dE}{dx} = F_T \cdot A^2 k^2 \sin^2 (\phi)$$

$$\left\langle \frac{dE}{dx} \right\rangle= F_T \cdot A^2 k^2 \left\langle \sin^2 (\phi) \right\rangle$$

$$\left\langle \frac{dE}{dx} \right\rangle = \frac{1}{2} F_T \cdot A^2 k^2$$

Then I simply considered that $k = \frac{2 \pi}{\lambda}$ and replaced the variables with the values the problem had already given me or I found out. The answer I got was $\left\langle \frac{dE}{dx} \right\rangle \approx 0.99$ J/m, which agrred with the official answer. However, I'm bothered with the step in which I had made $\frac{\partial y}{\partial t} = \frac{\partial y}{\partial x} \cdot \frac{d x}{d t}$. Is this chain rule correct or did I just got lucky? If it is indeed correct, was I correct in interpreting $\frac{d x}{d t}$ as the same thing as $v$? The argument I used to convince myself of it was considering a level curve $L$ of $y(x,t)$ so that $y(x,t) = y_0, \forall (x,t) \in L$. Then one may isolate $x$ in terms of $t$, take the derivative and see that $\frac{d x}{d t} = v$. Is this argument valid in the situation presented in this exercise?


In this specific case, it is possible to use this relation, although not necessarily due to the chain rule.

It is possible to analyze the situation in a discrete way. Assume a given point in the string in moving on the $Oy$ axis with velocity $u$ and on the $Ox$ axis with velocity $v$. The ratio between a displacement in the $Oy$ axis $\Delta y$ and a displacement in the $Ox$ axis $\Delta x$ is the same as the ratio between $u$ and $v$ (you can just divide by the time interval in both the numerator and the denominator). Therefore, we have

$$\frac{\Delta y}{\Delta x} = \frac{u}{v}$$ $$\frac{\Delta y}{\Delta x} = \frac{\left( \frac{\Delta y}{\Delta t} \right)}{\left( \frac{\Delta x}{\Delta t} \right)}$$ $$\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}$$

Taking the limit as $t \to 0$, we may write: $$\frac{\partial y}{\partial t} = \frac{\partial y}{\partial x} \cdot \frac{d x}{d t}$$


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