next up previous contents
Next: Motion in the x Up: General Analysis Previous: Identify the degrees of

Derive the equation of motion

To derive the equation of motion, we express Newton's law of motion for each particle (string segment) solely in terms of the solution qy(x,t) and constants specified in the problem. We therefore consider the free-body diagram for a single individual segment, which we draw in Figure 2 for the segment of string between positions x and tex2html_wrap_inline859 .

   figure82
Figure 2: Free-body diagram for a segment of string

Only two forces act on a given string segment, the tensions from the segments neighboring to the left and eight: no significant long-range forces actgif and the only other contact with the string is with the surrounding airgif. These tension forces act along the direction tangent to the string. Anticipating coming developments, the figure brakes the tension forces into x and y components. The tensions on either side of the segment need not be equalgif, and so we further identify the components of the tension as being measured either at point x ( tex2html_wrap_inline871 and tex2html_wrap_inline873 ) or at point tex2html_wrap_inline859 ( tex2html_wrap_inline877 and tex2html_wrap_inline879 ).

Newton's Law for the states

  equation93

where we use the version of Newton's law which applies to finite bodies so that we need not assume that the segment is a small point. In this version of Newton's law, we need only consider external forces acting on the system, namely the tensions acting on either end, tex2html_wrap_inline881 is the acceleration of the center of mass of the segment, and tex2html_wrap_inline883 is the mass of the entire segment. Because the mass per unit length of the string is tex2html_wrap_inline861 and the length of the segment is tex2html_wrap_inline887 gif, we have tex2html_wrap_inline891 .





Tomas Arias
Mon Oct 15 16:15:07 EDT 2001