# Positive oscillatory solution

1. Sep 7, 2011

### kraigandrews

1. The problem statement, all variables and given/known data
For the oscillatory solution (+k2) solution, suppose x is restricted to 0$\leq$x$\leq$a by requiring $\psi$(x)=0 and $\psi$(a)=0, what are the restrictions on A,B and k?

2. Relevant equations
given $\psi$(x)=Acos(kx)+Bsin(kx)

3. The attempt at a solution
so, I pretty just plugged in the zero values to get A must be zero, leaving only $\psi$(a)=0=Bsin(kx), which implies that B=0 and thus k can be 0 or pi, which seems to be a completely unreasonable result, so I am confused as to where I went wrong? Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 7, 2011

### LCKurtz

You have 0 = sin(ka) not sin(kx). So ka can be any integer multiple of pi: k = nπ/a.