Mohammad Asadzadeh - An Introduction to the Finite Element Method for Differential Equations

Consider a perfectly elastic and flexible string stretched along the segment of the ‐axis, moving perpendicular to its equilibrium position. Let denote the density of the string in the equilibrium position and the density at time . In an arbitrary small interval , the mass will satisfy, see Figure 1.4.

(1.5.15)

Figure 1.4A vibrating string.

Thus, using Lemma 1.1, ( 1.5.15) gives the conservation of mass:

(1.5.16)

Now we use the tensions and , at the endpoints of an element of the string and determine the forces acting on the small interval . Since we assumed that the string moves only vertically, the forces in the horizontal direction should be in balance: i.e.

(1.5.17)

Dividing ( 1.5.17) by and letting , we thus obtain

(1.5.18)

Hence,

(1.5.19)

where because it is the magnitude of the horizontal component of the tension.

On the other hand, the vertical motion is determined by the fact that the time rate of change of linear momentum is given by the sum of the forces acting in the vertical direction. Hence, using ( 1.5.16), the momentum of the small element is given by

(1.5.20)

with the time rate of change:

(1.5.21)

There are two kinds of forces acting on the segment of the string: (i) the forces due to tension that keep the string taut and whose horizontal components are in balance, and (ii) the forces acting along the whole length of the string, such as weight. Thus, using ( 1.5.19), the net tension force acting on the ends of the string element is

(1.5.22)

Further, the weight of the string acting downward is

(1.5.23)

Next, for an external load, with density , acting on the string (e.g. when a violin string is bowed), we have

(1.5.24)

Finally, one should model the friction forces acting on the string segment. We shall assume a linear law of friction of the form:

(1.5.25)

Now applying Newton's second law yields

(1.5.26)

Dividing ( 1.5.26) by and letting , we obtain the equation

(1.5.27)