Simply Armonic Movement and k constant

[Plat00n]

I have a question on "k" of an armonic simple movement.

If we take the equation of a wave, this is:

$$\frac {\partial \psi (x,t)}{\partial x^2} = \frac 1 v^2 \frac {\partial^2 \psi (x,t)}{\partial t^2}$$

And this, if I'm not wrong, must to satisfy the independent of time Helmholtz equation:

$$\frac {d^2 A(x)}{dt} + k A(x) = 0$$

I'm ok since here?

If it's ok, the solution of the first equation could be:

$$\psi (x,t) = A e^{kx-wt}$$

Is this "k" the same that the "k" in Helmholzt equation? Is there any mistake in this?

Plat00n.

 

[Plat00n]

I have a question on "k" of an armonic simple movement.
If we take the equation of a wave, this is:
$$\frac {\partial \psi (x,t)}{\partial x^2} = \frac 1 v^2 \frac {\partial^2 \psi (x,t)}{\partial t^2}$$
And this, if I'm not wrong, must to satisfy the independent of time Helmholtz equation:
$$\frac {d^2 A(x)}{dt} + k A(x) = 0$$
I'm ok since here?
If it's ok, the solution of the first equation could be:
$$\psi (x,t) = A e^{kx-wt}$$
Is this "k" the same that the "k" in Helmholzt equation? Is there any mistake in this?
Plat00n.

Nobody can help me a little?

 

[quantumdude]

I have a question on "k" of an armonic simple movement.
If we take the equation of a wave, this is:
$$\frac {\partial \psi (x,t)}{\partial x^2} = \frac 1 v^2 \frac {\partial^2 \psi (x,t)}{\partial t^2}$$
And this, if I'm not wrong, must to satisfy the independent of time Helmholtz equation:
$$\frac {d^2 A(x)}{dt} + k A(x) = 0$$
I'm ok since here?

Not quite. It should be:

$$\frac {d^2 A(x)}{dx^2} + k^2 A(x) = 0$$

If it's ok, the solution of the first equation could be:
$$\psi (x,t) = A e^{kx-wt}$$
Is this "k" the same that the "k" in Helmholzt equation? Is there any mistake in this?

If you use the corrected version of the equation that I posted, then yes the $k$'s are the same.