Wave Interference Simulation

1. My measurments for low amplitude was a wavelengh of 3.12 cm and when I changed it to a higher amplitude, my wavelengh droped to 1.86 cm

2. My measurments for a lower frequency is a wavelengh of 5.18 cm and when I changed to a higher frequency, my wavelengh changed to 1.12 cm

3. From this data I figured out that during a low amplitude and a high frequency, the wavelengh was smaller, but when I tested a high amplitude and a low frequency, the wavelengh became larger.

4. Drop 1 wavelength: 1.35 cm
Drop 2 wavelength: 1.10 cm


First measurments will be from drop one to each point (A,B,C,D,E,F):
A=3.14 cm  B=3.17 cm  C=5.29 cm  D=5.45 cm  E=5.55 cm  F=7.55 cm

Second measurments are from drop two (same points):
A=5.55cm  B=3.27 cm  C=3.14 cm  D= 7.75 cm  E=5.60 cm  F=5.50 cm

From gathering this information, I learned that from each "point", the distances would are similar if you were to think of it as looking into a mirror. What I mean by this is if you look at the measurements for drop one, then if you look at drop two, you will see that if you switch point A with C and point D with F, those measurements are similar in lengh.
Now how would we compair this to the distances between each of the points reletive in wavelengths? I would answer this by first finding out how long one wavelength would be. Take for example, if I made one wavelength equal to the length from point 1 (or 2) to point B, we would find that the wavelength equals  about 3.22 cm (taking the average distance). Now by using the wavelength in the example (3.22 cm), we can say that from pt. 1 to pt. C, the distance would be roughly 1.65 wavelenghs. The same could be said about pt. 2 to pt. A. If we continue with this, the wavelengh between pt. 1 (or 2) to pt. E would equal roughly 1.75 wavelengths. Lastly if we compair pt. 1 to pt. F (or pt. 2 with pt. D) the length would equal about 2.375 wavelengths.