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So the length of a wave is the distance between two compressed regions as shown in this representation of a longitudinal wave:

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But how do you know exactly where the two points are? Is there a point where the particles are most compressed? It's hard for me to explain my confusion with this. Are you measuring the distance between only two particles or are you taking the average of an area of many compressed particles and saying "somewhere in here is one point" and then doing the same for the other compressed region? Hope this question makes sense.

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You don't need a particular point on the wave. You only have to make sure it's the same point on each successive wave. If you have a microphone, hooked to an oscilloscope, you can measure the time between peaks (or troughs, or zero-crossings), multiply by the speed of sound, and that's your wavelength.

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    $\begingroup$ Not only that, but practically speaking you usually don't just measure 1 wave. You measure 10 or 100 waves and divide. Then if you measure from not exactly the right spot on the wave the error in estimating the wavelength won't be very large. $\endgroup$
    – The Photon
    Jul 2, 2015 at 15:19
  • $\begingroup$ "you can measure the time between peaks (or troughs, or zero-crossings)". Ok zero-crossings seems more concrete to me. But how do you know what point to measure in a peak or trough- would that just be half the distance between two zero crossings? $\endgroup$
    – red888
    Jul 3, 2015 at 12:18
  • $\begingroup$ @red888: Up to you. It's not too hard to slide a vertical line cursor to the peak of a curve so it looks, by eye, to be in the center of a peak - and it's the place where a tangent to the curve is horizontal. $\endgroup$ Jul 3, 2015 at 13:52
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So the length of a wave is the distance between two compressed regions as shown in this representation of a longitudinal wave:

is in general not true but rather it is a pictorial representation for simple cases in one dimension. A wave is any solution of a wave equation of the form $\Box \psi(\textbf{x},t) = 0$ that can be expressed in the form $$ \psi(\textbf{x},t)=\int d^3k\,d\omega\,\tilde{c}(\textbf{k},\omega)\,\textrm{e}^{-i(\textbf{k}\cdot\textbf{x} - \omega t)} $$ where the wavelength of each component $\tilde{c}(\textbf{k})=2\pi/\textbf{k}$. If you evaluate this in the simple case of just one Fourier component this reduces to taking the value of the function at any point $\textbf{x}$ and calculating the spatial distance from the next point $\textbf{x}'$ fulfilling $\psi(\textbf{x},t)=\psi(\textbf{x}',t')$, with $t'-t$ being the period of the wave.

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