# Why is $f=v/\lambda$ true for a standing wave? [closed]

If a wave travels at speed $v$ and has wavelength $\lambda$ it's obvious that any given place in the medium through which the wave is travelling will undergo one full oscillation every $v/\lambda$ units of time, and so the frequency $f$ of the wave must be given by $f=v/\lambda$.

But I can't see why a standing wave should obey that equation. What does it even mean to talk about the "speed" of a standing wave? I'd love an explanation.

(I'm asking this because I'm trying to understand the harmonic series of a plucked or struck vibrating string with fixed ends. Assume the length of the string is $1/2$. Then I understand that the string can only have waves whose wavelength is an element of $\{1,\ 1/2,\ 1/3, ...\}$. Assume that the wave with wavelength $1$ vibrates at frequency $f_!$. Then we can supposedly use the equation $f=v/\lambda$ to find that the wave with wavelength $1/2$ must vibrate at frequency $f_2=2*f_1$, the one with wavelength $1/3$ with frequency $f_3=3*f_1$ and so on, giving the harmonic series $f_1,f_2,f_3,...$, where $f_n=n*f_1$but as you can see from the above, I can't see why the $f_n$ should have the values they do.)

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A standing wave is just a superposition of two waves travelling in opposite directions. The relation applies to the two superimposed waves individually. – Julien Dec 29 '12 at 16:50

## closed as off topic by Nameless, Henry T. Horton, Davide Giraudo, Alexander Gruber, TMMDec 29 '12 at 18:04

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## 1 Answer

The fact that the higher modes of a string have frequencies that are integer multiples of the fundamental is not automatic in the properties of higher modes. It comes from the calculation of potential and kinetic energy in the string. Once you calculate these, the spacing of the modes comes out. In other oscillators, such as 2D drum heads or the semi-classical hydrogen atom, the frequencies have a different spacing.

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