# In a sound wave, is there rarefactions at both ends?

I was reading my book Physics, For scientists and Engineers, Third Edition, by Randall D. Knight, studying the first chapter on waves. This diagram is provided on page 565: The diagram is easy to follow, but I don't understand why the line of the graph goes below the abscissa (ie, there is rarefication in FRONT of the wave.)

I thought that a wave worked like Newtons cradle: one particle hits the particle in front of it, causing it to move - momentum is transferred through the medium (forcing the particles slightly closer together in the process) , and it's effects can be felt when some of this momentum is lost to physical motion of some other mass (I would assume different from the medium) - for example, it hits the cilia in your ears.

However if I'm right, the reduction in front of the wave would imply that somehow, as the first ball hits the second ball in Newton's cradle, the third ball is in turn attracted to the second!

I doubt this can be - so, why is there reduced pressure in front of the wave?

• I think that's just the form of that wave -- imagine it was a string in your hand and you moved it down, then up, then returned to zero. Contrast this with moving up then back to zero, which results in a Gaussian-like pulse (no rarefaction on either side). – tpg2114 Sep 29 '15 at 1:30

The graph shows the displacement of matter with respect to the equilibrium position. Since some matter is displacement toward the left in the lower graph (where the arrows go down left), this means that the displacement is in the opposite direction with respect to the direction of propagation. This is why the displacement is negative and thus the reason why the displacement $\Delta x$ goes below the axis $\Delta x=0$.