Trying to find wave equation for a string by abusing notation (algebraically manipulating differentials)

Question

I'm trying to derive the wave equation for a string of mass density $\mu$ with an equal tension $T$ being applied at both ends at an angle $\theta_x$ relative to the $x$-axis. I've taken an infinitesimal length of the string $dx$ and found the mass along this length to be $m=\mu dx$. I define the vertical displacement of the string to be $\psi$, then applying Newton's 2nd law I find:

$$\Sigma F_\psi = \mu dx \frac{d^2\psi}{dt^2}.$$

Since equal tension is applied on both ends, I have:

$$Tsin(\theta_{x+dx})-Tsin(\theta_x)=\mu dx \frac{d^2\psi}{dt^2} \implies \frac{sin(\theta_{x+dx})-sin(\theta_x)}{dx}=\frac{\mu}{T}\frac{d^2\psi}{dt^2}.$$

The tangent of the angle $\theta_x$ basically gives me the slope of the string: $tan(\theta_x)=\frac{\psi}{x}$, and sine is basically tangent for small angles. So I arrive at:

$$\frac{tan(\theta_{x+dx})-tan(\theta_x)}{dx}=\frac{\mu}{T}\frac{d^2\psi}{dt^2} \implies \frac{(\frac{d\psi}{dx})_{x+dx}-(\frac{d\psi}{dx})_x}{dx}=\frac{\mu}{T}\frac{d^2\psi}{dt^2}.$$

And finally I arrive at:

$$\frac{d}{dx}\left[\left(\frac{d\psi}{dx}\right)_{x+dx}-\left(\frac{d\psi}{dx}\right)_{x}\right]\stackrel{?}{=}\frac{\mu}{T}\frac{d^2\psi}{dt^2}.$$

And I do not know how to proceed from here and arrive at a wave equation in the form

$$u_{xx}=c^2u_{tt}.$$

I can see that I am close, the right hand side is already in that form, but I'm not sure how to algebraically manipulate the left hand side to be in the form I want. This derivation is found on page 14 of Anthony Gerig's "Introduction to Wave Physics". He appears to skip the step of manipulating the left-hand side, and it does not look trivial to me.

Answer 1

You have one notational issue that seems to be tripping you up. Your last equation should be $$ \frac{1}{dx} \left[ \left(\frac{d\psi}{dx}\right)_{x+dx}-\left(\frac{d\psi}{dx}\right)_{x}\right] = \frac{\mu}{T} \frac{d^2 \psi}{dt^2}. $$ (Note the lack of the "$d$" in the numerator out front.)

The quantity in square brackets can be recognized as the infinitesimal change in $d\psi/dx$ when we go from $x$ to $x + dx$, i.e., $$ \left(\frac{d\psi}{dx}\right)_{x+dx}-\left(\frac{d\psi}{dx}\right)_{x} = d \left( \frac{d\psi}{dx} \right) $$ which then implies that $$ \frac{1}{dx} \left[ \left(\frac{d\psi}{dx}\right)_{x+dx}-\left(\frac{d\psi}{dx}\right)_{x}\right] = \frac{d}{dx} \left[ \frac{d\psi}{dx} \right] = \frac{d^2 \psi}{dx^2} $$ as desired.

Answer 2

Your derivation is on the right path, but you have an "extra" derivative. Namely the dynamical equation of a small piece of the wire $\Delta x$, reads

$$\begin{aligned} \Delta x \mu \partial_{tt} w(x,t) & = - N \theta(x,t) + N \theta(x+\Delta x,t)+ f(x,t) \Delta x \\ \Delta x \mu \partial_{tt} w(x,t) & = N ( - \partial_x w(x,t) + \partial_x w(x+\Delta x,t) )+ f(x,t) \Delta x = \\ & = N ( - \partial_x w(x,t) + \partial_x w(x) + \Delta x \partial_{xx} w(x,t) + o(\Delta x)) + f(x,t) \Delta x= \\ & = \Delta x N \partial_{xx} w(x,t) + f(x,t) \Delta x \end{aligned}$$

and thus

$$\mu \partial_{tt} w - N \partial_{xx} w(x,t) = f(x,t) \ .$$