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Differential Wave Equation
Consider
a progressive transverse wave moving along an infinitively long rope, the wave
is generated by moving one end of the rope up and down with certain amplitude. Since the rope is
infinitively long, the back reflection of the wave at the other end of the rope
can be neglected and thus no standing wave effect in this case.
Figure 3.1 below illustrates the displacement of a part AB
of the rope with length ds from its
equilibrium position (AoBo,
with length dx). The tensile force
exerting on the string is F, for
simplicity, we don’t take the gravitational effect into account. ρ is the mass of the rope per unit
length.
Figure
3.1. A string (tensile force F) shows a
laterally deformed shape AB. The
equilibrium situation corresponds to the axial position y=0 (AoBo).
We are going to derive one-dimensional wavefunction for a
small amplitude. Applying
Newton’s law ΣF = ma for the forces
in y-direction, i.e. second
derivative of y with respect to time t is the acceleration due to displacement in y-direction, we obtain:
$F\sin (\alpha +d\alpha )-F\sin (\alpha
)=\rho ds\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$ (3.1)
Working out Equation (3.1), with: $\sin (\alpha +d\alpha
)=\sin (\alpha )\cos (d\alpha )+\cos (\alpha )\sin (d\alpha )$ , and for a
small wave amplitude, α is very small:
$\cos (d\alpha )\approx 1$ , $\sin (d\alpha )\approx d\alpha $, $\cos
(\alpha )\approx 1$ , and $dx\approx ds$, we then obtain:
$Fd\alpha =\rho dx\frac{{{\partial
}^{2}}y}{\partial {{t}^{2}}}$ (3.2)
Since $\tan \alpha =\frac{dy}{dx}$ and by differentiating
to x, we obtain:
$\frac{1}{{{\cos }^{2}}\alpha }\frac{d\alpha }{dx}=\frac{{{d}^{2}}y}{d{{x}^{2}}}$ . Thus,
$d\alpha ={{\cos }^{2}}\alpha dx\frac{{{d}^{2}}y}{d{{x}^{2}}}$ and Equation (3.2) can be rewritten:
$\frac{1}{{{\cos }^{2}}\alpha }\frac{d\alpha }{dx}=\frac{{{d}^{2}}y}{d{{x}^{2}}}$ . Thus,
$d\alpha ={{\cos }^{2}}\alpha dx\frac{{{d}^{2}}y}{d{{x}^{2}}}$ and Equation (3.2) can be rewritten:
$F{{\cos }^{3}}\alpha
\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=\rho \frac{{{\partial
}^{2}}y}{\partial {{t}^{2}}}$ (3.3)
For a small amplitude, ${{\cos }^{3}}\alpha \approx 1$ ,
and Equation (3.3) can be simplified to:
$\frac{{{\partial
}^{2}}y}{\partial {{x}^{2}}}=\frac{\rho }{F}\frac{{{\partial }^{2}}y}{\partial
{{t}^{2}}}$ (3.4)
The constant term $\frac{\rho }{F}$ is usually expressed
in term of $\frac{1}{{{c}^{2}}}$ (i.e. ${{c}^{2}}=\frac{F}{\rho }$ ), and c has the dimension of velocity, the so
called phase velocity of the wave. The phase velocity gives the speed at which
the wave profile propagates. Equation (3.4) can be rewritten as follows:
$\frac{{{\partial
}^{2}}y}{\partial {{x}^{2}}}=\frac{1}{{{c}^{2}}}\frac{{{\partial
}^{2}}y}{\partial {{t}^{2}}}$ (3.5)
This is thus the differential form of wave equation, and has a
general solution in this form:
$f(x,t)=g(x\pm ct)$ (3.6)
For waves propagate in + or – x-direction.
