Wave Equation (Fourier Coefficients)

[zack7]

For the wave equation I managed to get
the coefficient of f:

a₁=2

and

the coefficient of g:

$\frac{12pi}{2pi*2}$=B₂

Is these answers right, since my B₂ does not match the answer I was given.

Thank you

 

[haruspex]

Was that the right attachment? I don't see any reference to a1, B2, f or g there.

 

[zack7]

Was that the right attachment? I don't see any reference to a1, B2, f or g there.

Basically I was taught the the first U(x,0) is f and the second is g, so I go on to name the Fourier coefficient of f coefficient as a and the second as b.

 

[haruspex]

Ok. In future, please don't assume others use the same notation.
I get b₂ = 2. Please post your working.

 

[zack7]

Ok. In future, please don't assume others use the same notation.
I get b₂ = 2. Please post your working.

Using the formula ($\frac{n*pi*a}{l}$)*b₂=12

where a=2, ;L=pi, and n=2,

Therefore I get three, I think I am making a mistake with my a.

 

[haruspex]

But I don't know where you get that formula from. What equations do you have for u(x,t) and u_t(x,t) before determining b2?

 

[zack7]

That formula was given by my teacher after forming the formula for u(x,t)

A_n= Fourier Coefficient of f

($\frac{n*pi*a}{l}$)*b₂=Fourier Coefficient of g

The u(x,t) formula is

$\sum$(A_n*cos$\frac{npiat}{l}$+B_n*sin$\frac{npiat}{l}$)*sin$\frac{npix}{L}$

where a infinite series is not needed for this question

 

[haruspex]

I understand that you get it down to a₁ cos(3t) sin (x) + b₂ sin(6t) sin(2x), right? And we agree that the initial condition on u gives a₁=2. But show me how you use the condition on u_t to get b₂ =3.

 

[zack7]

I understand that you get it down to a₁ cos(3t) sin (x) + b₂ sin(6t) sin(2x), right? And we agree that the initial condition on u gives a₁=2. But show me how you use the condition on u_t to get b₂ =3.

Looking at the function 12sin(2x)

I get n= 2
L= pi (initial condition)
Fourier coefficient = 12
a= 2 ( previous answer (a₁)

Using the formula I get
$\frac{12pi}{2pi*2}$=B₂
which gives me three which is wrong since the answer should be two

 

[haruspex]

zack, please please please show me all your working.
You have u(x, t) = a₁ cos(3t) sin (x) + b₂ sin(6t) sin(2x)
You are given u(x, 0) = 2 sin(x), from which you deduce a₁=2.
You are given u_t(x, 0) = 12 sin(2x).
Show me what equation you get for u_t(x, 0), and how setting that equal to 12 sin(2x) gives you b₂=3.
You refer to some other equation you have been given, but I don't know exactly what that equation is or in what context you are supposed to apply it. My guess is you are misusing it.

 

[zack7]

zack, please please please show me all your working.
You have u(x, t) = a₁ cos(3t) sin (x) + b₂ sin(6t) sin(2x)
You are given u(x, 0) = 2 sin(x), from which you deduce a₁=2.
You are given u_t(x, 0) = 12 sin(2x).
Show me what equation you get for u_t(x, 0), and how setting that equal to 12 sin(2x) gives you b₂=3.
You refer to some other equation you have been given, but I don't know exactly what that equation is or in what context you are supposed to apply it. My guess is you are misusing it.

This was the equation that my teacher used when he found B_n , I do not know any other way to find it.

($\frac{n*pi*a}{l}$)*b₂=Fourier Coefficient of g

 

[haruspex]

This was the equation that my teacher used when he found B_n , I do not know any other way to find it.

Fourier coefficients are determined by plugging in the initial condition information. There is no generic formula. I would guess the formula your teacher gave you was for a certain kind of initial condition and does not apply here. Please try my way. If you can't follow it all at first, post what you can.
- Write out the equation for u_t(x, t)
(You understand that this means the partial derivative of u(x,t) wrt t, right?)
- Plug in the given fact that u_t(x, 0) = 12 sin(2x).
- From that, determine b₂.
In view of your misunderstanding on this point, it might also be useful for you to redo the step where you determine a₁. This time, instead of using any formula your teacher gave you, work it from the other initial condition you were given.

 

[zack7]

Fourier coefficients are determined by plugging in the initial condition information. There is no generic formula. I would guess the formula your teacher gave you was for a certain kind of initial condition and does not apply here. Please try my way. If you can't follow it all at first, post what you can.
- Write out the equation for u_t(x, t)
(You understand that this means the partial derivative of u(x,t) wrt t, right?)
- Plug in the given fact that u_t(x, 0) = 12 sin(2x).
- From that, determine b₂.
In view of your misunderstanding on this point, it might also be useful for you to redo the step where you determine a₁. This time, instead of using any formula your teacher gave you, work it from the other initial condition you were given.

Using this formula $\frac{2}{L}$$\int(f(x)*sin(\frac{npix}{L}$)

I obtain A_n= 2

and

b_n= 12

what should I do next ?

Is this the way you were talking about, now I would need to find B_n

 

[haruspex]

Using this formula $\frac{2}{L}$$\int(f(x)*sin(\frac{npix}{L}$)

I obtain A_n= 2

and

b_n= 12

zack, we're never going to get through this if you keep writing things like "I obtain". Show me precisely how you obtain it. Leave nothing out. If you quote a formula, explain your understanding of what all the terms in the formula mean.

 

[zack7]

zack, we're never going to get through this if you keep writing things like "I obtain". Show me precisely how you obtain it. Leave nothing out. If you quote a formula, explain your understanding of what all the terms in the formula mean.

Using this formula $\frac{2}{L}$$\int(f(x)*sin(\frac{npix}{L}$)

for f(x) = 2sin(x)
n=1 since cx=x which gives me one
L=pi from the conditions of the wave equation where x goes from 0 to pi

This gives me
$\frac{2}{pi}$$\int(2sin(x)*sin(\frac{1pix}{pi}$)

$\frac{2}{pi}$$\int(2sin(x)*sin(x)$)

$\frac{2}{pi}*2$$\int(\frac{1}{2}-\frac{1}{2}cos(2x))$)

This gives me
x-sin(x)cos(x)
Inserting the limits gives me pi*$\frac{2}{pi}$
gives me 2

Therefore A₁=2

I did the same way for b_n by solving the integral which gives me 12

 

[haruspex]

Let me try a different tack. You understand that the secondary boundary condition refers to u_t(), not u()? How do you use that when you apply your formula? Are you using a different formula for getting b₂ from that you used to get a₁?

 

[zack7]

Let me try a different tack. You understand that the secondary boundary condition refers to u_t(), not u()? How do you use that when you apply your formula? Are you using a different formula for getting b₂ from that you used to get a₁?

I used the same boundary condition for the integral below (0<x<pi)
$\frac{2}{pi}$$\int(12sin(2x)*sin(\frac{2pix}{pi}$)

This gave me 12

Is this wrong ?

 

[haruspex]

I used the same boundary condition for the integral below (0<x<pi)
$\frac{2}{pi}$$\int(12sin(2x)*sin(\frac{2pix}{pi}$)

This gave me 12

Is this wrong ?

I would think so. It is not going to be exactly the same procedure for u() and for u_t(). I don't know what formula to give you instead - not an expert on Fourier analysis - but I can solve it very easily from first principles. You have a generic equation for u, so you can differentiate wrt t (partially) to get u_t. Then substitute in the initial condition info.

 

[zack7]

Thank you for all the help, I got it solved, got my teacher to explain.