Consider the general displacement of a particular point on a
string. Let the coordinates of the point, the equilibrium point of
one rope atom, be x,y,z. Then we use the symbol 
to represent the displacement of the rope atom from its equilibrium
position. The displacement is, of course, different for each atom (each
atom) so that the displacement is a function of the atomic position, i.e. 
. Moreover, this displacement
changes with time, so that we really have 
. This is a vector displacement from an equilibrium
position. As with any vector, we can write it in terms of its components,
i.e.
Now, to simplify this complicated expression.
1.Let us take the string to be along the z axis, so that every point has only one coordinate, namely z. Both the xand the y coordinates are zero. Therefore, we have
2. Motion along the z direction is longitudinal motion and motion along
either x or y is transverse motion. I will ignore the possibility of
longitudinal motion. This means there is no possible displacement in the z
direction, or that 
. Thus we now have
3. Linear Polarization. This means that, while there are two possible
transverse directions and, hence, any possible vector combination of these
two motions, we will take all of the transverse motion to be solely along
one axis (which I call the x axis). There is, thus, only one transverse
direction of motion,. Since this motion can only be along 
, this is
linear motion (straight line motion). We then have
But, since we no longer really have a vector, we can just write
as the displacement.
Now, let us consider the forces acting on some small piece of string, whose
length will be 
and whose mass will be 
. Moreover, I
will let the string be uniform, that is, any length 
will have the
same mass 
no matter what the z value is. Then I can define a
constant number, called the linear mass density,
such that 
.
Now we will look at the diagram below to get the restoring force on some
small length 
.
The restoring force in the x direction is clearly given by
We hope to express the force 
in terms of 
and its spatial derivative. We know that
Rewrite the force as
But we now have the tangent which is a derivative so we get
To further simplify this, we consider a function
and use our knowledge of Taylor Expansions to write the following:
The quantity 
If the quantity 
we will neglect all quadratic and higher order
terms in 
. Then
Notice that the subscript 1 has been dropped. I have done this because 
is small and, therefore, the point within 
where we take
the derivative is not really important. Any point will do!
Finally, therefore, we have the restoring force
and this is equal to mass times acceleration. Since acceleration is the second derivative with respect to time we get, then
Since 
we get finally
or
The term
has the units of inverse velocity squared, as we see from
Therefore, we arrive at the WAVE EQUATION
We have already seen this equation for electromagnetic radiation where 
was replaced by E and v by the speed of light c. Same equation,
incredibly different system (but only in appearance to us, not in its
physical behavior!).