A sigle wave
Superposition of waves
© The scientific sentence. 2010
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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.
Next:
. 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.
©: The scientificsentence.net. 2007.
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