Standing wave with specific initial/boundary conditions

[Isow]

Homework Statement

"Solve the wave equation with the following initial conditions and boundary conditions."

∂²Y/∂x² = 1/v² * ∂²Y/∂t²

Boundary conditions:
∂Y/∂x(x=0, t)=0 and Y(x=L,t)=0

Initial Conditions:
∂Y/∂t(x, t=0) = 0
Y(x,t=0) = δ(x-L/2)

Homework Equations

Using separation of variables:

Y(x,t) = X(x)*T(t)

X(x) = C*cos(kx/L) + D*sin(kx/L)
T(t) = E*cos(wt) + F*sin(wt)

The Attempt at a Solution

So, I'm relatively familiar with this process but I have a couple of hangups relating to these specific initial conditions. First, the requirement that the partial of y with respect to x=0 at all times t (first boundary condition) seems, to me, to make a wave impossible. It gives D=0 AND C=0 OR k=0, but of course all of these are unsatisfactory solutions.

More analytically, it seems that requiring the slope to always be 0 at x=0 is really restricting that piece of the wave from moving. I'm just not sure how to handle it.

Moving past that though (I assumed y(x=0,t)=0 as my first boundary condition giving C=0), I'm also not sure how to handle the initial condition. The initial condition equation, x-L/2, has an integral of 0 when integrated from 0 to L. This makes sense since the expression has equal area above and below y=0, but I still don't know how to handle it in terms of getting an actual solution.

Just thinking out loud as I type this, does the delta in the expression indicate that I should be treating it as an absolute value equation so that all of it is forced to be above y=0?

 

[SammyS]

Homework Statement

"Solve the wave equation with the following initial conditions and boundary conditions."

∂²Y/∂x² = 1/v² * ∂²Y/∂t²

Boundary conditions:
∂Y/∂x(x=0, t)=0 and Y(x=L,t)=0

Initial Conditions:
∂Y/∂t(x, t=0) = 0
Y(x,t=0) = δ(x-L/2)

Homework Equations

Using separation of variables:

Y(x,t) = X(x)*T(t)

X(x) = C*cos(kx/L) + D*sin(kx/L)
T(t) = E*cos(wt) + F*sin(wt)

The Attempt at a Solution

So, I'm relatively familiar with this process but I have a couple of hangups relating to these specific initial conditions. First, the requirement that the partial of y with respect to x=0 at all times t (first boundary condition) seems, to me, to make a wave impossible. It gives D=0 AND C=0 OR k=0, but of course all of these are unsatisfactory solutions.

More analytically, it seems that requiring the slope to always be 0 at x=0 is really restricting that piece of the wave from moving. I'm just not sure how to handle it.

Moving past that though (I assumed y(x=0,t)=0 as my first boundary condition giving C=0), I'm also not sure how to handle the initial condition. The initial condition equation, x-L/2, has an integral of 0 when integrated from 0 to L. This makes sense since the expression has equal area above and below y=0, but I still don't know how to handle it in terms of getting an actual solution.

Just thinking out loud as I type this, does the delta in the expression indicate that I should be treating it as an absolute value equation so that all of it is forced to be above y=0?

The Boundary condition, ∂Y/∂x(x=0, t)=0, only makes D = 0 with no restriction on C.
The other Boundary condition gives you k .

It's a standing wave. Right?

What are the conditions at nodes and at anti-nodes?

 

[Isow]

Hey, thanks, that was exactly what I needed! I realized the first boundary condition only set D=0, but I erroneously was thinking that the second boundary condition gave C=0 when, in fact, you're right: it gives k.

I then applied the second initial condition in terms of absolute value, which I think is the correct way to go about it, and everything worked out.

Thanks again!

edit: and yes, it's a standing wave.

 

[SammyS]

Hey, thanks, that was exactly what I needed! I realized the first boundary condition only set D=0, but I erroneously was thinking that the second boundary condition gave C=0 when, in fact, you're right: it gives k.

I then applied the second initial condition in terms of absolute value, which I think is the correct way to go about it, and everything worked out.

Thanks again!

You're very welcome.