# Homework Help: Wave equation, taut string hit with hammer

1. Dec 10, 2017

### Taylor_1989

1. The problem statement, all variables and given/known data
A string of length L is fixed at both ends $u(0,L)=u(L,t)=0$ The string is struck in the middle with a hammer of width a, leading to an intial condtion $u(x,0)=0$ and

$$U_t(x,0)=v_0$$ for $$\frac{l}{2}-\frac{a}{2} \leq x \leq \frac{l}{2}+\frac{a}{2}$$

and

$$U_t(x,0)=0$$ other wise

I have printed screen in the question Just encase it make no sense what I have wirtten

I am not going to put full working, I am having trouble with one particular part the intial condtions for my Fourier series.

2. Relevant equations
$$U_t(x,0)=\sum_{n=1}^\infty \left(-\frac{n\pi c}{L}\right)B_n sin(\frac{n\pi}{L}x)$$ [1]

$$\left(-\frac{n\pi c}{L}\right)B_n=\frac{2}{l}\int _0^lv_0sin\left(\frac{n\pi }{L}x\right)dx\:$$ [2]

$$B_n=-\frac{2v_0}{-n\pi c}\int _{\left(\frac{L}{2}-\frac{a}{2}\right)}^{\frac{L}{2}+\frac{a}{2}}sin\left(\frac{n\pi }{L}x\right)dx\:$$ [3]
3. The attempt at a solution

[3] soultuion
$$B_n=\frac{2Lv_0}{n^2\pi ^2c}\left(cos\left(\frac{n\pi }{2}\left(\frac{L+a}{L}\right)-cos\left(\frac{n\pi }{2}\left(\frac{L-a}{L}\right)\right)\right)\right)$$

I belive I have gone wrong with the limits and my problem lies in how the size of the hammer is effecting the wave in this case, could someone please advise, thanks in advace

2. Dec 10, 2017

### Ray Vickson

The wave equation has a speed parameter "$v$" in it; why does your solution have no $v$ anywhere? Is your $c$ perhaps equal to the $v$ in the question?

3. Dec 10, 2017

### Taylor_1989

Sorry yes my c=v my appolgise. I have also just had a throught could I use the indenties for cos(a+b) and cos(a-b) to solve this problem

Last edited: Dec 10, 2017
4. Dec 10, 2017

### LCKurtz

I agree up to here except I don't have a minus sign. But that shouldn't affect the final answer.

That isn't quite right. Apparently the $l$ should be $L$, but the function isn't $v_0$ on that interval. So it is better written$$\left(-\frac{n\pi c}{L}\right)B_n=\frac{2}{L}\int _0^L u_t(x,0) \sin\left(\frac{n\pi }{L}x\right)dx$$ $$\left(-\frac{n\pi c}{L}\right)B_n=\frac{2}{L}\int _{\frac L 2 - \frac a 2} ^{\frac L 2 + \frac a 2 }v_0 \sin\left(\frac{n\pi }{L}x\right)dx$$

I didn't check your final answer but I don't see any obvious major errors.