Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $0$.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
thanks! you're a lifesaver –  rapidash Aug 13 '11 at 1:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.