I am trying to understand the wave equation of a string fixed on both the ends, which looks like this:

$$ \frac{\partial^2 y}{\partial x^2} - \frac1{c^2}\frac{\partial^2y}{\partial t^2} - \gamma\frac{\partial y}{\partial t}- l^2\frac{\partial^4y}{\partial x^4} = 0 $$

For the below equations, $\rho$ stands for linear density, $b$ for damping coefficient, $T$ for tension, and $c$ stands for the wave's speed.

I am comfortable in the case where I do not consider damping. I could bring this result:

$$ \frac{\partial^2 y}{\partial x^2} -\frac1{c^2}\frac{\partial^2y}{\partial t^2} = 0 $$


$$ F = T\cdot\sin(\theta + \Delta\theta) - T\cdot\sin (\theta) $$

and $$ F = \rho\Delta x \frac{\partial ^2y}{\partial t^2} $$

However, when I take damping into consideration and derive it this way:

$$ F = T\cdot\sin(\theta + \Delta\theta) - T\cdot\sin(\theta) - b\cdot \frac {\partial y}{\partial t} $$

$$ \rho\Delta x \cdot \frac{\partial^2 y}{\partial t^2} = T\cdot \sin (\theta +\Delta \theta) - T\cdot\sin(\theta) -b\cdot\frac{\partial y}{\partial t} $$

$$ \rho\Delta x \cdot \frac{\partial^2 y}{\partial t^2} = T\cdot \tan (\theta +\Delta \theta) - T\cdot\tan(\theta) -b\cdot\frac{\partial y}{\partial t} $$

$$ \frac{\partial^2 y}{\partial t^2} = \frac1{\rho\Delta x}\left(T\cdot\left(\frac{\partial y (x+\Delta x,t)}{\partial x} - \frac{\partial y(x,t)}{\partial x}\right) - b\cdot\frac{\partial y}{\partial t}\right) $$

$$ \frac{\partial ^2y}{\partial x^2}-\frac1{c^2}\frac{\partial^2y}{\partial t^2}-\frac\gamma{\Delta x}\frac{\partial y}{\partial t} = 0 $$

This is the problem. I have a $\Delta x$ term where I am not supposed to.

  • 3
    $\begingroup$ “$\rho$ stands for linear density, $b$ for damping coefficient”… There’s neither a $\rho$ nor a $b$ in your first equation. I strongly suggest you type the equations in MathJax format and this way you’ll also “manually enforce” consistency in the equations (which is not the case now). $\endgroup$
    – ZaellixA
    Sep 23, 2023 at 14:34
  • 3
    $\begingroup$ I strongly suggest you try to use the same symbols/letters in all equations… What is $\rho$ and $b$ in the first equation? Are they even present? Are they relevant? Again, I’ll suggest the use of MathJax for the equations. $\endgroup$
    – ZaellixA
    Sep 23, 2023 at 15:31
  • 2
    $\begingroup$ Question doesn't seem to have any relationship with the "fixed on both ends" constraint in the title or the boundary conditions tag $\endgroup$
    – Sten
    Sep 23, 2023 at 16:47
  • 1
    $\begingroup$ The $ \gamma \frac{\partial y}{\partial t}$ term applies external damping (like air resistance) to each position. I don't think this is accurate in terms of structural damping. $\endgroup$ Sep 23, 2023 at 16:55
  • $\begingroup$ I am not familiar with the $\frac{\partial^4 y}{\partial x^4}$ term for a string. I have only seen this with beams. $\endgroup$ Sep 23, 2023 at 20:20

1 Answer 1


$$F=T\sin (\theta+\Delta\theta)-T\sin\theta-b \frac{\partial y}{\partial t}$$

This is the wrong starting point. $F$ is the force on an arbitrarily small length $\Delta x$ of string. As $\Delta x$ goes to 0, the force has to go to 0. (Otherwise the acceleration would go to infinity!)

That constraint is correctly respected by the first two terms, which cancel in the limit that $\Delta x\to 0$, because then $\Delta \theta\to 0$ also (assuming the string is smooth). But the last term fails to satisfy this constraint.

  • $\begingroup$ Thanks! Quite silly of me to miss that. This means the starting point got to be this. $$ F = T\cdot\sin(\theta + \Delta\theta) - T\cdot\sin(\theta) - b\cdot \frac {\partial y}{\partial t}\Delta x;$$ But then the last term does not have the units of force. $\endgroup$ Sep 24, 2023 at 12:20
  • $\begingroup$ @AnanthaKrishnan If that term represents some sort of drag force, then presumably it should scale with the size of the segment under consideration. What is $b$? $\endgroup$
    – Sten
    Sep 24, 2023 at 19:22
  • $\begingroup$ b is damping coefficient (Newton-second per metre). Won't multiplying with delta x, lead to Newton-metre instead of having Newton alone? $\endgroup$ Sep 26, 2023 at 4:47
  • $\begingroup$ @AnanthaKrishnan I think you want the damping coefficient itself to scale proportionally to $\Delta x$ then. Maybe think about what is physically causing the damping and whether that should scale with size. (For example, air resistance would scale with size because a larger segment hits a larger volume of air.) $\endgroup$
    – Sten
    Sep 26, 2023 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.