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So I have to do this:

$$-\sum_{n=0}^{\infty} \left(\frac{n^2 \pi \ T}{L^2} a_n(t) + m_La^{''}_n\right)\sin\left(\frac{n \pi x}{L} \right) = F_0 \sin \omega_E t.$$

I'm supposed to multiply this equation through by $\sin(kx\pi/L)$, integrate over $x$ from $0$ to $L$, and use the fact that

$$\int_0^L \sin\left(\frac{n \pi x}{L} \right) \sin \left(\frac{k \pi x}{L} \right)dx = \begin{cases}\frac{L}{2}& \mbox{if }n=k\\ 0&\mbox{otherwise} \end{cases}$$

to derive a set of ordinary differential equations that model each individual $a_k(t)$.

I tried to do this and got stuck, so I thought I'd come here for help.

This is what I have so far

If $n = k$ $$-\sum_{n=0}^{\infty} \left ( \int_0^L { \left(\frac{n^2 \pi \ T}{L^2} a_n(t) + m_La^{''}_n\right)}dx \times \frac L 2 \right ) = \int_0^L {F_0 \sin \omega_E t \sin \left(\frac{k \pi x}{L} \right)}dx$$ $$-\sum_{n=0}^{\infty} \left(\frac{n^2 \pi \ T}{L} a_n(t) + Lm_La^{''}_n \right) \times \frac L 2 = \int_0^L {F_0 \sin \omega_E t \sin \left(\frac{k \pi x}{L} \right)}dx$$ $$-\sum_{n=0}^{\infty} \left( 2n^2 \pi \ T a_n(t) + 2L^2 m_La^{''}_n \right) = \int_0^L {F_0 \sin \omega_E t \sin \left(\frac{k \pi x}{L} \right)}dx$$ $$-\sum_{n=0}^{\infty} \left( 2n^2 \pi \ T a_n(t) + 2L^2 m_La^{''}_n \right) = F_0 \sin \omega_E t \left(- \frac{L}{k \pi}\right) \left(\cos \left(k \pi \right) - 1\right)$$ and then i have no idea. If $n \neq k$, then the right side is the same and the left side is $0$.

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up vote 1 down vote accepted

Is the sum from n=0 to infinity still? ;) Use the n /= k condition to drop that to a single term.

Edit: Thought I would be a bit more explicit.

$-\sum_{n=0}^\infty\Big(\frac{n^2\pi T}{L^2}a_n(t)+m_L\ddot{a_n}\Big)sin(\frac{n\pi x}{L})=F_o sin(\omega_E t)$

$\int_0^L \Big[-\sum_{n=0}^\infty\Big(\frac{n^2\pi T}{L^2}a_n(t)+m_L\ddot{a_n}\Big)sin(\frac{n\pi x}{L})\Big]sin(\frac{k\pi x}{L})=\int_0^LF_o sin(\omega_E t)sin(\frac{k\pi x}{L})$

Now look at each piece of the sum. For each one that is $n \neq k$, it is 0. For each term that is $n=k$, multiply by $\frac{L}{2}$. So:

$-\Big(\frac{k^2\pi T}{L^2}a_k(t)+m_L\ddot{a_k}\Big)\frac{L}{2}=\int_0^LF_o sin(\omega_E t)sin(\frac{k\pi x}{L})$

Notice the complete substitution from n to k.

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thanks! you're a lifesaver – rapidash Aug 13 '11 at 1:22

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