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A sigle wave

Superposition of waves

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Equation of the harmonic wave

The relationship (3.3): y(x,t) = A sin[(2π/λ)(x - vt)]
is the harmonic wave function. From it, we will find the equation of the harmonic wave.

Let's consider the partial derivate of this harmonic wave function:
. with respect to time t (holding x constant):
∂y/∂t = - v(2π/λ) A cos[(2π/λ)(x - vt)]
Wich gives the y-component of the velocity of an element.
The second derivative gives:
2y/∂t2 = - v2(2π/λ)2 A sin[(2π/λ)(x - vt)]
= - v2(2π/λ)2 y(x,t)      (4.1)
Wich gives the y-component of the acceration of an element.
. with respect to x (holding t constant):
∂y/∂x = A (2π/λ) cos[(2π/λ)(x - vt)]
This derivative gives the slope of the function at a time t and point x. The second derivative gives:
2y/∂x2 = - A (2π/λ)2 sin[(2π/λ)(x - vt)]
= - (2π/λ)2 y(x,t)      (4.2)
This second derivative gives the bending of the function. While x is increasing, as it is shown in the figure 3, where "a" is the acceleration ∂2y/∂x2:

  • If ∂2y/∂x2 > 0, the slope increases, the function bends upward and the acceleration is directed upward.
  • If ∂2y/∂x2 < 0, the slope decreases, the function bends downward and the acceleration is directed downward.
  • If ∂2y/∂x2 = 0, the slope is constant, the function is straight at that point and the acceleration is null.

Combining the equations (4.1) and (4.2), we have:
2y/∂t2 / [- v2(2π/λ)2] = ∂2y/∂x2/[- (2π/λ)2]
or: ∂2y/∂t2 / v2 = ∂2y/∂x2
Rearranging, we obtain:
2y/∂x2 - (1/v2)∂2y/∂t2 = 0      (3)

2y/∂x2 - (1/v2)∂2y/∂t2 = 0

This differential equation is the wave equation.
Since we have obtained this equation from the derivatives of the wave function; then the wave function satisfies the wave equation; and the solution of the wave equation are, from now on, the wave functions. More generaly, any f(x - vt) or f(x + vt) are also solution of the wave equation.

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