1
$\begingroup$

According to the wave equation for 1-D which is $$\frac{\partial ^2y}{\partial x^2}-\frac{1}{v^2}\frac{\partial^2y}{\partial t^2}=0$$ which can be derived from $y(x,t)=f(x\pm vt)$ using calculus.

Moreover, the wave equation above can also be derived by considering transverse waves on a string as well, using Newtonian mechanics and some small angles approximate.

What I am confused about is that if that transverse wave has large amplitude, the terms that have been canceled out using the approximation will remain in the equation and it won't satisfy the wave equation above, even though the wave with large amplitude can also be describe by this function ($y(x,t)=f(x\pm vt)$)

Could anyone please tell me the reason why? Thank you very much!

$\endgroup$

1 Answer 1

4
$\begingroup$

Your assumption that large amplitude transverse waves still satisfy $y(x,t) = f(x\pm v t)$ is not correct. If there are any additional nonlinear terms or higher order derivatives in the differential equation due to the breakdown of the linear assumption, $F[y]$, then the differential equation is

$$ \frac{\partial ^2y}{\partial x^2}-\frac{1}{v^2}\frac{\partial^2y}{\partial t^2} + F[y] =0 $$ which won't, in general, have d'Alembertian solutions.

A potentially relevant form for $F[y]$ is $F[y] = \frac{\kappa}{2} \left( \frac{\partial y}{\partial x} \right)^2$, which I've seen come up in the context of modeling elastics systems. In terms of real-world systems, like piano strings, coupling to longitudinal modes can be relevant as well.

$\endgroup$
3
  • $\begingroup$ Thank you very much, your answer is really helpful! And could you please explain more about why does large amplitude transverse wave won't satisfy the wave function (𝑦(𝑥,𝑡)=𝑓(𝑥±𝑣𝑡), Are there any resistance force added to the system when the amplitudes are bigger? $\endgroup$
    – RKJ
    Oct 29, 2021 at 19:18
  • 1
    $\begingroup$ It's just like with regular springs: for small displacements, the restoring force is basically linear in the amount of stretching, for large displacements there are non-linear terms. Another limit to think about is how can one model the case where a portion of the string becomes perpendicular, or even folds back, on itself. In those extreme cases the concept of "transverse displacement" itself breaks down, and you can't express the configuration of the string as a single valued function $y(x,t)$. $\endgroup$
    – Dave
    Oct 29, 2021 at 19:36
  • $\begingroup$ Thank you so much! $\endgroup$
    – RKJ
    Oct 30, 2021 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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