A sigle wave
Superposition of waves
© The scientific sentence. 2010

Harmonic wave equation and Newton's second law
Let's consider a string wiggled once in the (xy) plan from its equilibrium
position. The wave produced propagates along the + x axis using the string
as a medium. Un element of the string is represented as follows:
This element is bent because of the forces F_{1} and F_{2}
exerted at the two ends of this element by the two neighboring elements.
We assume that the disturbance is small, such that the force F called a
tension along the string is uniform (F = F_{1} = F_{2});
and the slope of the string is not large (the angles θ_{1} and θ_{2}
are small; that is the wavelength λ >> the amplitude A of the wave ).
The net force on the element is :
ΣF = F_{1} + F_{2} = (F_{1x} + F_{2x}, F_{1y} + F_{2y}) (1)
Its xcomponent is: F_{1x} + F_{2x} = F(cosθ_{2}  cosθ_{1}). As θ is
small, we can use the MacLaurin series cosθ = 1  (1/2) θ^{2} + ... and consider just the
first term wich is 1, to find that component is null.
It remains then the y component of the net force.
F_{1y} + F_{2y} = F(sinθ_{2}  sinθ_{1}) =
F(tanθ_{2}  tanθ_{1}) ( θ is small)
But tanθ is the slope of the function at the point x, that is ∂y/∂x. So
tanθ_{2}  tanθ_{1} = [∂y/∂x]_{2}  [∂y/∂x]_{1} =
Δ[∂y/∂x] = (Δ[∂y/∂x]/Δx) Δx = (∂^{2}y/∂x^{2}) Δx
Then:
F_{1y} + F_{2y} = F (∂^{2}y/∂x^{2}) Δx (2)
If the mass of the string is M and its length is L, the linear mass density μ ; that
is the mass per unit length is μ = M/L. The related mass of the element is m = μ Δ x.
The Newton's second law for the element is:
ΣF = m (∂^{2}y/∂t^{2})
It becomes:
F (∂^{2}y/∂x^{2}) Δx = μ Δ x (∂^{2}y/∂t^{2})
Thus:
(∂^{2}y/∂x^{2})  (μ/F) (∂^{2}y/∂t^{2}) = 0
(3)
Equating this equation with the wave equation (4.3), we obtain:
(μ/F) = 1/v^{2}. Then:
v = [F/μ]^{1/2} (4)
The speed of a wave depends on the mass of the medium where this
wave moves along.
©: The scientificsentence.net. 2007.


