1
$\begingroup$

Standing waves are the waves in which disturbances do not simply propagate forward or backward, but rather the material particles are moving up and down continuously, with the particles between two consecutive nodes in the same phase. But when we are talking about the harmonics of such a wave, we define the $n^{th}$ harmonic by- $$f_n=\dfrac{n}{2L} \sqrt{\dfrac{T}{\mu}}$$ The problem is, this formula is derived by assuming that the wave speed is $$v=f\lambda$$ or $$v=\sqrt{\dfrac{T}{\mu}}$$ But if the wave is actually standing, what does this speed actually mean? Does it mean twice the length of the string times the frequency of the individual particles performing harmonic oscillations? And if that is so, who is to say that it will be equal to $\sqrt{T/\mu}$?

$\endgroup$
1
  • $\begingroup$ Could you provide more details about the system/equation that you are solving, how you arrive at harmonics, and why you think you need to know the speed. The question seems to lack the context. $\endgroup$
    – Roger V.
    Sep 5, 2020 at 8:02

2 Answers 2

4
$\begingroup$

What makes a standing wave is having waves in a single medium propagating in opposite directions at exactly the same speed, and in phase. This requires the waves to possess a unique speed through the medium in order to establish the "standing wave" condition (and it also requires the medium to have a finite length into which an integer number of those waves can fit). That's why the standing wave condition must account for the wave speed.

$\endgroup$
1
$\begingroup$

Actually, in the standing waves theory this expression will add almost nothing to your study, it becomes really significant only when you deal with traveling waves. As @niels nielsen said, if you want, you can see too the standing waves as a superposition of traveling waves, but, as you can see, even seeing from this side requires the idea of ​​waves propagating. Too, this is the phase speed, that is, the speed which the phase propagates, but in standing waves the phase don't propagate, it just stay there!

$\endgroup$

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.