What were the steps taken to get from point A to point B in this forced vibrations problem? I'm working through the derivation of the forced response (vibration) of a cantilevered beam. I have a basic understanding of the derivation until this point.
screenshot of derivation
I can see how the summation(s) are equivalent to an integral - but I don't see why $q_n$ and/or $\omega$ wouldn't be included within that integral, since q is dependent on $n$. I can rationalize how to get the same result if $\omega$ and $q$ are regarded as constants (not dependent on $n$) but that doesn't seem to be the case (?).
I have a feeling the "orthogonality conditions" they mention have something to do with it, but I haven't been able to discern how from my research. In general, I have limited knowledge of linear algebra, and I haven't had to use vector calculus concepts of like orthogonality since taking the course a couple years ago. Any help is appreciated :)
Full document (screenshot is from pgs. 28 + 29): http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf
 A: Supposing the orthogonality conditions are
$$
\int_0^l W_m(x)W_n(x)dx=\delta_{mn}b.
$$
Now, multiplying (84) by $W_m(x)$
$$
\sum_{n=1}^\infty \omega_n^2W_m(x)W_n(x)q_n(t)+\sum_{n=1}^\infty W_m(x)W_n(x)\frac{dq_n^2(t)}{dt}=\frac{f(x,t)}{\rho A}W_m(x)
$$
then integrate over $x\in[0,l]$
$$
\sum_{n=1}^\infty \omega_n^2q_n(t)\int_0^lW_m(x)W_n(x)dx+\sum_{n=1}^\infty \frac{dq_n^2(t)}{dt}\int_0^lW_m(x)W_n(x)dx=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx
$$
then using orthogonality
$$
\sum_{n=1}^\infty \omega_n^2q_n(t)\delta_{mn}b+\sum_{n=1}^\infty \frac{dq_n^2(t)}{dt}\delta_{mn}b=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx
$$
or, considering the Kronecker delta
$$
\omega_m^2q_m(t)b+\frac{dq_m^2(t)}{dt}b=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx
$$
dividing by $b$
$$
\omega_m^2q_m(t)+\frac{dq_m^2(t)}{dt}=\frac{1}{\rho A b}\int_0^lf(x,t)W_m(x)dx
$$
and using the definition of $Q_n(t)$
$$
\omega_m^2q_m(t)+\frac{dq_m^2(t)}{dt}=\frac{1}{\rho A b}Q_m(t)
$$
Given the arbitrarity of the index $m$ (this relation holds for each $m$), you can rename the index $n$ and you have (85).
