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Force Waves

Figure C.14: Transverse force propagation in the ideal string.

Referring to Fig.C.14, at an arbitrary point $ x$ along the string, the vertical force applied at time $ t$ to the portion of string to the left of position $ x$ by the portion of string to the right of position $ x$ is given by

$\displaystyle f_{rl}(t,x) = K\sin(\theta) \approx K\tan(\theta) = Ky'(t,x)$ (C.41)

assuming $ \left\vert y'(t,x)\right\vert \ll 1$, as is assumed in the derivation of the wave equation. Similarly, the force applied by the portion to the left of position $ x$ to the portion to the right is given by

$\displaystyle f_{lr}(t,x) = - K\sin(\theta) \approx - Ky'(t,x).$ (C.42)

These forces must cancel since a nonzero net force on a massless point would produce infinite acceleration. I.e., we must have $ f_{rl} +
f_{lr} \equiv 0$ at all times $ t$ and positions $ x$.

Vertical force waves propagate along the string like any other transverse wave variable (since they are just slope waves multiplied by tension $ K$). We may choose either $ f_{rl}$ or $ f_{lr}$ as the string force wave variable, one being the negative of the other. It turns out that to make the description for vibrating strings look the same as that for air columns, we have to pick $ f_{lr}$, the one that acts to the right. This makes sense intuitively when one considers longitudinal pressure waves in an acoustic tube: a compression wave traveling to the right in the tube pushes the air in front of it and thus acts to the right. We therefore define the force wave variable to be

$\displaystyle f(t,x) \isdefs f_{lr}(t,x) = -Ky'(t,x).$ (C.43)

Note that a negative slope pulls up on the segment to the right. At this point, we have not yet considered a traveling-wave decomposition.

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