# Changing the wave equation

1. Mar 27, 2006

### MarkB

What happens when you make the x varible in the wave egn to some power

m(dx/dt) + k x^n =0

What happens when n increases/decreases?

2. Mar 27, 2006

### daveb

Um, perhaps you might want to get the correct form of the wave equation first?

3. Mar 27, 2006

### benorin

That diff. Eq. is seperable, and has the solution

$$x(t)=\left( C_1+(n-1)\frac{k}{m}t\right) ^{-\frac{1}{n-1}}$$

which is vastly different from

m(dx/dt) + k x =0

which is also seperable, and has the solution

$$x(t)=C_2e^{-\frac{k}{m}t}$$

but it is notable that the limit as n->1 of the former solution is the later solution if $$C_1=C_2=1$$.

Last edited: Mar 27, 2006
4. Mar 27, 2006

### nrqed

I guess you meant C times (1+ (n-1).....)

It's interesting. It's neat to see the exponential recovered as n ->1.

5. Mar 27, 2006

### HallsofIvy

Staff Emeritus
If that were a second derivative, then you would be closer to a wave equation.

6. Mar 27, 2006

### MarkB

Yeah sorry I forgot to make the eqn: m(d^(2)x/dt^(2)) + k X^(n)=0

this is the eqn of the wave cause by a simple harmonic Isolator.