# 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 the...water, parallel to the line joining the sources....

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 the...

water, parallel to the line joining the sources. The first nodal lines (the ones adjacent to the central axis) cross the metre stick at the 35.0 cm and 55.0 cm marks. Each of the crossing points is 50.0 cm from the midpoint of the line joining the two sources. Calculate the wavelenth and speed of the waves.

Frequency stays in Hz, while cm to m.

unkyd | High School Teacher | (Level 2) Adjunct Educator

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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