1-D wave equation with fixed ends and initial displacement 2sin(πx)

[ColdFusion85]

It is a 1-D wave equation problem with fixed ends, no initial velocity, and initial displacement of $$2sin(\pi x)$$ on the interval 0<x<4, t>0.

See my attached documents of my work. I end up with a c_n value of 0 based on the integration. I am pretty confident I set up the problem correctly as it follows a similar, generalized problem in my class notes. Obviously I used actual numbers in this problem. Can anyone spot any errors in my work? Thanks for your help.

 

[ColdFusion85]

Anyone?

 

[montoyas7940]

Have you tried to read the thumbnails, ColdFusion85. The quality is poor.

 

[ColdFusion85]

No they aren't. Place your cursor over the picture, the magnifying glass appears. Just click and you will be able to see it fine.

 

[happyg1]

unless you have a mac and then you use SHIFT +

 

[ColdFusion85]

Looking for help again. Can anyone figure out what's wrong?

 

[Gear300]

Try moving this thread to the section Differential Equations...looks like that's what this is. You'll probably have a higher chance of getting help there.

 

[Kreizhn]

At the end of your first page, knowing that $\sqrt{\lambda} = \frac{n \pi}{4}$ you went from
$$T^{\prime \prime} - 9 \lambda T = 0 \text{ to } T^{\prime \prime} + \frac{9 n^2 \pi ^2}{16} T = 0$$.
Why'd you switch sign?

 

[HallsofIvy]

You have
$$\sum C_n Sin(\frac{n\pi x}{4})= 2 Sin(\pi x)$$
for the initial value and then start calculating a complicated Fourier coefficient.

Isn't it obvious that if C₄= 2, all other C_n= 0 satisfies that? Since the sines are "orthogonal", your integral should have been 0 for all n except n= 4. You don't need the full sine series when your function is a single sine!