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I seem to remember watching a video where someone took a rope which was fixed at one end and gave it a jerk (or maybe more than one), and then did something to the end he was holding (fixed it somehow), so the pulse he created resulted in a displacement of the rope which was constant. This was explained as being the result of the waves bouncing back and forth between the fixed ends, and somehow interfering constructively to result in a time-constant displacement of the rope.

I've looked online and can't find the video, and my memory is hazy enough for me to distrust it on this matter. I am wondering if what I am describing is even possible, and if so, under what conditions. More formally:

The general solution to the wave equation in one dimension was given by d'Alembert as $$u(x,t) = f_1(x-ct) + f_2(x+ct),$$ where $f_1$ models disturbance traveling in the positive $x$ direction and $f_2$ in the negative $x$ direction. Is there any selection of initial conditions $u(x,0)$ and $u_t(x,0)$ ($u_t$ is the partial of $u$ w.r.t. time $t$) as well as boundary conditions for the string (e.g. $u(0,t) = 0$ for all $t$), and selection of $f_1, f_2$ such, for some $\tau \geq 0$ and any real number $r$ we have

$$u(x,\tau) = u(x,\tau+r) \neq 0~?$$

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  • $\begingroup$ Are you talking about standing waves? en.wikipedia.org/wiki/Standing_wave $\endgroup$ – FrodCube Dec 31 '16 at 15:51
  • $\begingroup$ That makes me strongly think to the stroboscopic effect, wasn't that? $\endgroup$ – user130529 Dec 31 '16 at 16:03
  • $\begingroup$ Standing waves on a string do not have time-constant displacement, so no, not that. And it wasn't the stroboscopic effect either (at least not in my memory), though that's a clever response. $\endgroup$ – Wapiti Dec 31 '16 at 16:17
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    $\begingroup$ I know an effect which accomplishes this behaviour: the end isn't fixed though, instead a rope is moved upwards at speeds close to the speed of propagation. This causes the rope to appear stationary/very slow. This effect can be seen in this video: youtube.com/watch?v=BSSecjKRxEE $\endgroup$ – AccidentalTaylorExpansion Dec 31 '16 at 16:50
  • $\begingroup$ Not stationary, very slow. That is cool though. $\endgroup$ – Wapiti Jan 3 '17 at 19:27
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No, there is not. Far from the boundary, boundary conditions do not matter so that the only problem is fixing a function $f=f(x)$ which solves D'Alembert equation. It reduces to $f''(x)=0$ with the general solution $f(x) = ax+b$. Since I think you are looking for a disturbance confined in a region between the endpoints, the only solution of that type is the trivial one $f(x)=0$ everywhere.

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  • $\begingroup$ I follow and appreciate your reasoning, but does this really establish the impossibility? You have not included time as a variable, and the only way that what I am describing could be possible is if two (or more) time varying solutions with nonzero acceleration combined to have zero acceleration but nonzero displacement. $\endgroup$ – Wapiti Dec 31 '16 at 16:38
  • $\begingroup$ For $t > \tau$ my reasoning applies, so the shape of the static wave must be $ax+b$ for some $a,b \in \mathbb R$. Actually your $\tau$ does not exist for this reason. If $g=g(t,x)$ is a solution of D'Alembert equation, for every fixed $t_0$, the pair $g(t_0,x), \partial_t g(t_0,x)$ uniquely determine the solution both in the future and the past of $t_0$. Take $t_0>\tau$, the function $g(t,x) = ax+b $ satisfies both D'Alembert equation and initial conditions. So that function is THE solution also in the past and also before $\tau$. $\endgroup$ – Valter Moretti Dec 31 '16 at 16:56
  • $\begingroup$ In other words there is no way to fix initial conditions at $t_1$ in order to have a static solution from $t_2>t_1$ on which is not static between $t_1$ and $t_2$. $\endgroup$ – Valter Moretti Dec 31 '16 at 17:00
  • $\begingroup$ I haven't accepted because I'm still not convinced. What about the Fourier transform of the indicator function, $f(x) = 1$ for $-1 \leq x \leq 1$ and 0 otherwise? This exists and gives a continuous spectrum of sinusoids which, added together, create a nonzero displacement with zero acceleration. What am I missing? $\endgroup$ – Wapiti Jan 8 '17 at 15:34
  • $\begingroup$ Are these sinusoids also function of time? Fourier (inverse) transform is not a superposition of elementary solutions of D'Alembert equation. $\endgroup$ – Valter Moretti Jan 8 '17 at 15:46

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