# Infinite string attached to a spring [closed]

I'm trying to solve a problem related to waves on a string.

Say I have an infinite string, with tension $$T$$ and mass density $$\mu$$.

To the string, at $$x=0$$ (seeing as it's infinite, the specific point doesn't actually matter, so long as it's constant), I attach a spring $$k$$ (the string is horizontal whilst the spring is vertical).

I am attempting to calculate the reflected and transmitted waves, resulting from an incident wave: $$y_{inc}(x,t)=A_{inc}\,e^{i(x-\omega\,t)}$$

My attempt at a solution was to write

$$\mu\,\frac{\partial^2y}{\partial t^2}|_{x=0}=T\,\frac{\partial^2y}{\partial x^2}|_{x=0}-k\,y(0,t)$$

which (I think) is, generally speaking, true. I am not sure how (if at all) I can use this to find an expression for the reflected wave.

So, my question is how does the spring affect the reflection?

• If I'd try to solve it, my first attempt would be to impose boundary conditions. In case the other end of this spring is not fixed, I expect the problem to be much more complicated than otherwise. Commented Nov 22, 2017 at 0:26
• Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. Commented Nov 22, 2017 at 11:04
• @KyleKanos this isn't homework. I'm trying to build this system and measure some of these things, but I had difficulty solving the analytical part. Commented Nov 22, 2017 at 11:06
• @ItamarG3: please check out the links I provided for a discussion on how we define homework/homework-like problems and how to post such questions. Commented Nov 22, 2017 at 11:07
• @KyleKanos no problem :) Commented Nov 22, 2017 at 11:10

There must be a body attached to the spring. Otherwise with $m = 0$ ideal spring, the motion of the string could never strech the spring (due to no 'inertia') and the problem would be equivalent to the one without the spring.
Say I attach a body of mass $m$ to the spring and let $s_0(t)$ describe its position relative to $y = 0$ string equilibrium position. $$m \ddot{s_0}(t) + k(-y(0,t) + s_0(t) - l_0) = 0$$ where $l_0$ is the equilibrium length of the spring. It's easier to use relative displacements: $s(t) = s_0(t) - l_0$ so $$m \ddot{s}(t) + k s(t) = ky(0,t). \tag1$$ Resultant force on the point $x = 0$ on the spring is 0: $$k(s(t) - y(0,t)) + T (\tan{\alpha_+} - \tan{\alpha_-}) = 0 \tag2$$ where $\tan{\alpha_\pm} = \partial_x y(0^\pm,t)$ Seek the solution of the wave-equation in the form \begin{align} y_{\leftarrow}(x,t) &= e^{i (\omega t - \bar k x)} + R e^{i (\omega t + \bar k x)}\\ \tag{As} y_{\rightarrow}(x,t) &= \Theta\, e^{i (\omega t - \bar k x)}. \end{align} where $R, \Theta \in \mathbb{C}$. Writing $s(t) = S e^{i \omega t}$ and using the $(\text{As})$ ansatz for (1) and (2) and imposing $$1 + R = \Theta \tag3$$ continuity gives three equations for three unknowns $(S,R,\Theta)$ in terms of $(m,\, \bar k,\, \omega = \bar{k}\sqrt{\frac{T}{\mu}}, \,T, \,\mu)$. An interesting fact is that for a configuration there can exist a wavenumber $\bar k_*$ for which the system isn't invertible. What is the physical explaination for it?