Info

-urn

Before evaluating this integral it should be noted that the square of a sum (as appearing in the integrand) can be written in terms of a double sum. This can be demonstrated by the following simple example:

= «1 (fli + a2 + 03) + «2 («1 + «2 + 03) + «3 («1 + «2 + «3)

j=1 /=i /=i j=1 Thus, the potential energy becomes j* 00 00 /> ^

For the string, the mode shapes and their first derivatives are sinusoidal functions; consequently, they form an orthogonal set.2 That is,

Thus, the potential energy relation can be simplified to j V «t

The integral in this expression can be integrated by parts as ¡•1 1 pi

/ 4>'ttfidx = 4>'t4>i - <pi(x)(p"(x)dx. (2.97)

Jo 0 Jo

By virtue of the boundary conditions at both ends, the first term is zero. Substitution of Eq. (2.40) into the last term (i.e., the integral) shows that

T (p'2dx — may2 / </>?</*= M,-£w?, (2.98)

Jo Jo

2 It is »of true in general that the derivatives of mode shape functions form an orthogonal set.

where M, was previously defined as the generalized mass and (», as the natural frequency of the i th mode. (It is noted that the /th generalized mass depends on the mode shape of the /th mode.) Thus, the potential energy becomes

0 0