Judging from the graph, it looks like it behaves as a periodic waves. It's got similar patterns that repeat 4 times. The 5th one is cut short, so it's not taken in the data analysis.
The probe took a little bit to process the data, but it was data that was acceptable to analyze.
The period of this waves is defined as the time it takes to measure one cycle. We can take the difference in times between cycles, and then divide it by the actual number of cycles that are covered in that time lapse.
T = (0.0271±.00005 - 0.0023±.00005)/4 = 0.00620 ± 0.000018
s
Having found the period of a wave, we can find the frequency of a sound wave through the relationship:
f = 1/T = 1/(0.00620 ± 0.000018) = 161.3 ± .47 s^-1
We can calculate the wave length of these waves in a medium (ie. air at v = 340 m/s) with the expression; v = f λ
λ = v/f = 340 / 0.00620 ± 0.000018 = 2.108 ± 0.0061 m. This is about the length of one of the whiteboard in front of the classroom.
The amplitude of the wave can be determined graphically by taking the average pressure of the highest and lowest points.
A = (2.712 ± .0005 - 2.576 ± .0005)/2 = 2.6440 ± 0.00036 m
If the data took 10 times longer to record, the only thing that would change is just the time. If the intensity was the same, the amplitude should not change. However, since the frequency is dependent of period, and the period is dependent of time and number of cycles, frequency, wavelength and period would be affected.
A second test was run by having a different person speak through the mic.
The following graph was acquired:
For this sample, we determined it would be better to use just 3 waves.
We came up with a period of;
T = (0.0250±.00005 - 0.0003±.00005) / 3 = 0.00823±.000024 s
and a frequency of;
f = 1/T = 121.5 ± .35 s^-1
The wave length of this sample was;
λ = v/f = 340 / 121.5 ± .35 = 2.798 ± 0.0081 m
Once we were done with taking vocal samples, we proceeded to experiment with tuning forks.
For the first tuning fork trial we got this graph:
We took into account only the first 10 cycles.
We found the period to be;
T = (0.0251±.00005 - 0.0016±.00005) / 10 = 0.002350 ± 0.0000050 s
with a frequency of;
f = 425.5 ± 0.91 s^-1
The wavelength came up to be,
λ = 0.799 ± 0.0017 m.
We ran a second trial with the same fork. This time, however, we hit the fork with a softer surface (rubber). This is the graph that we came up with:
This graph yielded virtually the same results for most of the data from the first fork trial, with a period of T = 0.002350 ± 0.0000050 s for 10 waves, a frequency f of f = 425.5 ± 0.91 s^-1, and therefore a wavelength λ = 0.799 ± 0.0017 m. One thing changed, and that was the amplitude that looked shorter than the first trial. The amplitude for this run was A = 2.711m.
