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


1 Answer 1


Your first equation is applying a force at all points $0<x<L$ 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 $$ u(x=0,t)= A \sin \omega t. $$

  • $\begingroup$ That's what I tried as I say in Edit2, but it also fails to give me an unique normal mode as a solution, probably that's because in this scenario the string is not fixed at both sides anymore, and we don't get the normal modes expected $\endgroup$ Feb 16, 2020 at 14:33
  • 1
    $\begingroup$ 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? $\endgroup$
    – mike stone
    Feb 16, 2020 at 16:06
  • $\begingroup$ As you say, I get that the therm with n=2 blows up to infinity, in a real world scenario, with damping, maybe this amplitude is really really high, so the motion is basically this normal mode. Is this what I saw happening in the lab? $\endgroup$ Feb 17, 2020 at 0:48

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