In phase by a common oscillator implies that the 2 waves are the
same. Since they are opposing, this implies that they are symmetric
(about an anti-node...i think...draw it out) which implies that the
number of nodes will be even, hence the 7th position will be NONE.
Now, the position of the anti-node (the axis of symmetry or
whatever ppl call it) is 1.23m/2 = .615 m
Now, use frequency 900 Hz to solve for wavelength = .38111
Now, we know that the space between a node and an adjacent
anti-node is equal to wavelength/4.
So, starting from the middle anti-node (@ .615m) and
adding/subtracting wavelength/4 will lead you to the positions of
the 3rd/4th nodes.
Note that since the nodes are symmetric, we can just find the 3
nodes (on one side of the symmetry axis) and subtract the positions
of those nodes from 1.23 m ( the total distance) to get the other 3
nodes (on the other side of the symmetry axis).
.615 - .3811/4 = .5197 , this is the position of the 3rd
and 1.23 - .5197 = .7103 is the position of the 4th node
For the 2nd/5th nodes, use the rule that the space between two
adjacent nodes is wavelength/__?__ (hehehe) and just add/subtract
to/from the 3rd/4th node.
.5197 - wavelength/___?___ = .32915 is the 2nd node
and 1.23 - .32915 = _____ is the 5th node.
You can find the 1st/6th nodes similarly.