Two point sources, 5.0 cm apart, are operating in phase, with a common frequency of 6.0 Hz, in a ripple tank. A metre stick is placed above...

This is similar to Young's experiment where an
interference pattern is produced on a screen a distance (L) from the sources. The
problem I believe is that Young's equation works with some assumptions like the distance
between the sources (2.5 cm) needs to be much much smaller than the distance to the
screen (50 cm)

When I used Young's equation I
got:

wavelength = d*y/(n*L) = 2.5 cm * 10 cm /(.5 * 50 cm)
= 1 cm

where:

d = distance
between sources

y = distance between central antinode to
the node or antinode being evaluated

n = order of the node
or antinode (how many wavelength further from one source to y than to the other source
to y)

L = distance of screen to
sources

If this, were the case then speed = wavelength *
frequency

= 1 cm * 6.0 hz = 6.0
cm/s

or

= .01 m * 6.0 hz = .06
m/s

I looked further into the geometry of the problem and
found the following:

The diagram you describe creates an
isosceles triangle with side lengths of 50 cm and a base length of 20 cm (35 cm mark to
50 cm mark).  The altitude of this triangle (the central axis to the midpoint of the
sources which must intersect the meterstick at the 45 cm mark) can be calculated using
Pythagorean theorem.

50^2 = 10^2 +
altitude^2

altitude = sqrt(2500 - 100) = 48.989
cm

Shifting this Vertex (located at the midpoint between
the sources) to the right source makes the isosceles triangle into a scalene triangle
with the left side becoming the longer side (LS) and the right side becoming the shorter
side (SS) and the base remaining at 20 cm.  The segment that connects the 45 cm mark to
the right source (CS) can be calculated using Pythagorean theorem
again.

2.5^2 + 48.989^ =
CS^2

CS = 49.052 cm

The angle
CS would make with the original altitude (Q) can be
found:

sinQ = 2.5/49.052  => Q = arcsin(2.5/49.052)
= 2.92 degrees

so CS makes an angle of 92.92 degrees with
the left side of the meterstick and 87.08 degrees with the right
side.

Using the Law of Cosines you can find
LS:

LS^2 = 10^2 + 49.052^2 - 2*10*49.052
cos(92.92)

LS = 50.558
cm

Using Law of Cosines you can also find
SS:

SS^2 = 10^2 + 49.052^2 - 2*10*49.052
cos(87.08)

SS = 49.559 cm

The
process would be the same if the Vertex was moved to the source on the left.  Just the
short side (49.559 cm) would be on the left and the long side (50.558 cm) would be on
the right.

When a node is created the waves are meeting
exactly out of phase.  The first nodal line is produced becasue the wave has to travel
exactly one-half of a wavelength further from the far source compared to the distance a
wave from the near source must travel.  In this case the node at the 35 cm mark on the
meterstic is formed because the wave had to travel 50.558 cm from the source on the
right while it traveled 49.559 cm from the source on the left.  This is a difference of
 .999 cm or 1 cm.  So if half of a wavelength is 1cm then one wavelength is 2
cm.

If this is the case (I think more likely) then
everything doubles.

So

speed =
wavelength * frequency

=2 cm * 6.0 hz = 12.0
cm/s

or

= .02 m * 6.0 hz = .12
m/s