The one-dimensional waves They are those that propagate in a single direction regardless of whether the vibration occurs in the same direction of propagation or not. A good example of them is the wave that travels through a taut string like that of a guitar..
In a flat wave cross, the particles vibrate in a vertical direction (they go up and down, see the red arrow in figure 1), but it is one-dimensional because the disturbance travels in only one direction, following the yellow arrow.
One-dimensional waves appear quite frequently in everyday life. The following section describes some examples of them and also of waves that are not one-dimensional, to clearly establish the differences.
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Here are some examples of one-dimensional waves that can be easily observed:
- A pulse of sound that travels through a straight bar, since it is a disturbance that propagates the entire length of the bar.
- A wave that travels through a channel of water, even though the displacement of the water surface is not parallel to the channel.
- Waves that propagate on a surface or through three-dimensional space can also be one-dimensional, as long as their wave fronts are planes parallel to each other and travel in only one direction..
An example of a non-one-dimensional wave is found in waves that form on a still water surface when a stone is dropped. It is a two-dimensional wave with a cylindrical wavefront.
Another example of a non-one-dimensional wave is the sound wave that a firecracker generates by exploding at a certain height. This is a three-dimensional wave with spherical wave fronts.
The most general way to express a one-dimensional wave that propagates without attenuation in the positive direction of the axis x and with speed v is, mathematically:
y (x, t) = f (x - v.t)
In this expression Y represents the disturbance in the position x Instantly t. The shape of the wave is given by the function F. For example the wave function shown in figure 1 is: y (x, t) = cos (x - v t) and the wave image corresponds to the instant t = 0.
A wave like this, described by a cosine or sine function, is called harmonic wave. Although it is not the only waveform that exists, it is of utmost importance, because any other wave can be represented as a superposition or sum of harmonic waves. It is about the known Fourier theorem, so used to describe signals of all kinds.
When the wave travels in the negative direction of the x-axis, it simply changes v for -v in argument, leaving:
y (x, t) = g (x + v t)
Figure 3 shows the animation of a wave traveling to the left: it is a shape called a function lorentziana and her mathematical expression is:
y (x, t) = 1 / (1 + (x + 1⋅t)two
In this example the speed of propagation is v = 1, -one unit of space for each unit of time-.
The wave equation is a partial derivative equation, the solution of which is of course a wave. It establishes the mathematical relationship between the spatial part and the temporal part of it, and has the form:
The following is the general expression y (x, t) for a harmonic wave:
y (x, t) = A⋅cos (k⋅x ± ω⋅t + θo)
a) Describe the physical meaning of the parameters A, k, ω Y θo.
b) What is the meaning of the ± signs in the argument of the cosine?
c) Verify that the given expression is indeed the solution of the wave equation from the previous section and find the velocity v propagation.
The characteristics of the wave are found in the following parameters:
-TO represents the amplitude or "wave height".
-what is in wave number and is related to the wavelength λ through k = 2π / λ.
-ω is the fangular frequency and is related to the period T wave oscillation by
ω = 2π / T.
-θo is the initial phase, which is related to the starting point of the wave.
A negative sign is taken if the wave travels in the positive direction of the X axis and a positive sign otherwise..
Verify that the given expression is a solution of the wave equation is simple: the partial derivative of the function is taken y (x, t) with respect to x twice, partially re-derivative with respect to t twice, and then combine both results to obtain an equality:
Second derivative with respect to x: ∂twoy / ∂xtwo= -Ktwo. TO⋅cos (k⋅x ± ω⋅t + θo)
Second derivative with respect to t: ∂twoy / ∂ttwo= -Ωtwo. TO⋅cos (k⋅x ± ω⋅t + θo)
These results are substituted in the wave equation:
-ktwo. TO⋅cos (k⋅x ± ω⋅t + θo) = (1 / vtwo) (-ωtwo. TO⋅cos (k⋅x ± ω⋅t + θo))
So much TO as the cosine are simplified, since they appear on both sides of the equality and the argument of the cosine is the same, therefore the expression reduces to:
-ktwo = (1 / vtwo) (-ωtwo)
Which allows to obtain an equation for v in terms of ω Y k:
vtwo = ωtwo / ktwo
v = ± ω / k
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