# Finding the conditions for a resonant (unique single mode) state in the wave equation

If we have a string of length $$L$$ in constant tension $$T$$ and we oscillate one of the extremes at the rate $$\sin{\frac{\pi nc t}{L}}$$, we observe in a lab that single harmonics are produced. I want to theoretically check that answer using the wave equation $$u_{tt}-c^{2}u_{xx}=\sin{\frac{n \pi c t }{L}}.$$ (The boundary conditions are for a fixed string with initial position and velocity $$0$$.)

In which the term in the right represents this sinusoidal force. However when I solve the PDE for a certain n ($$n=2$$ for example), I expected all the other modes to vanish, so I get a single normal mode that oscillates at that frequency $$\frac{2 \pi c t}{L}$$. However, I don't get this physically observed answer. What I am doing wrong? Which conditions should I set in the wave equations to check what I saw in the lab?

Edit: I'm not interested in the PDE solving, I'm interested in the physical conditions that, once set in the wave equation, produce the observed result

Edit2: I also tried solving the homogeneous wave equation stating that $$u(L,t)=\sin{\frac{n \pi c t}{L}}$$ but didn't get the expected result either

Edit3: I think the first method fails because I am applying it to all the points in the string at once, and the second one fails because in the lab we were told to considerate the two extremes as fixed

Your first equation is applying a force at all points $$0 of the string rather than at one end. It therefore couples to all of modes with $$n$$ an odd number. You will get some motion in all of these modes from this equation.
You force does not couple to the $$n=2$$ mode $$u(x,t)= \sin(2\pi x/L) \sin(2\pi ct/L)$$ because $$\int_0^L \sin (2\pi x/L) \sin(2\pi ct/L)dx=0.$$ If you have solved correctly you should get no motion from your equation in this case .
If you want to mimic the experimental condition you will need a different equation--- one that will depend on exactly how you apply the external force. If you shake the end transvesely, you can keep $$u_{tt}-c^2 u_{xx}=0$$ but solve with boundary condition $$u(x=0,t)= A \sin \omega t.$$
• For any shaking you will get some excitation of the non-resonant modes. In the absence of damping and at $\omega= 2\pi n/L$ the resonant mode will have infinite amplitude, so you need to stay away from exact resonance. What do you actually get when you solve? – mike stone Feb 16 at 16:06