## J

To prove that the mode shapes obtained for the uniform string problem are orthogonal, an individual modal contribution given by vf-(jc,i) = 0/W?/(O (2.36)

is substituted into the governing differential equation (wave equation) to obtain d2Vi d2Vi

Because the generalized coordinate is a simple harmonic function for the homogeneous solution,

Hi - -<»] Hi, (2.38) and the wave equation becomes

so that

If this procedure is repeated by substituting the j th modal contribution into the wave equation, a similar result,

is obtained. After multiplying Eq. (2.40) by </>, and Eq. (2.41) by </>;, subtracting, and integrating the result over the length of the string, we obtain

(wj - CO2) m [ v)0/( v)(/v = T f WiWkx) - cP"(.x)cPj(x)] dx. (2.42) J 0 J 0

The integral on the right-hand side can be integrated by parts using fb

J a a *> a by letting u — </);, du — fydx, v — <p'j, dv — (p"dx for the first term and u — <j)j, du — (p'jdx, v — 4>'j, dv — <p"dx for the second term. The result becomes

- T(<t>i<t>'j - </>;</>/)Io -T f Wtfj - 4>'i4>'j)dx - 0. J o

Note that the right-hand side is zero because </>,• is zero at both ends by virtue of the original boundary conditions. It may now be concluded that &>, / oj, when i ^ j, so that

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