# 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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