Data Data Table 1: Trial Values at Varying Degrees Length of String 0.025 25.5 g

Question

Data Data Table 1: Trial Values at Varying Degrees Length of String 0.025 25.5 g = 1 kg Avg. Time Period 5 cycles 100 cm = 1 m Mass of Bob Placemnt of Amplitude Trial 1Trial 2 bob Degrees (bob horizontal displacement) cm Trial 3 (s) 5 cycles 1 cycle 5 cycles5 cycles 9.91 10.13 10.19 10.12 10.22 10.08 0.38 10.23333333 .94 10.11333333 9.84 10.03333333 9.78 10.01 60 69 81 86 94 10.09 76 10.0710.33 10.16 9.84 10.1 9.75 9.94 9.75 10 Data Table 2: Trial values for bob masses Length of string cm Im litude 10 Bob weight (g) Bob weight TriTrial 2 Trial 3 (s) Avg Time (s) Period (kg) 50 9.68 9.04 9.62 9.446666667 Table 3: Trial values for string length Amplitude 10 Mass of bob g Trial 2 Trial 3 Avg Time (s) Length (m Trial 1 (s) Period 4.5 4.53 6.85 8.62 9.69 4.5 6.91 8.6 9.88 4.51 6.92 8.596666667 9.753333333 0.25 8.57 9.69 0.75

Explanation / Answer

from the given data

time period can be calculated by knowing the time period for n ossclations, and dividing it by n

so for the first chart

Time period

placement of bob time period

5 2.016 s

10 2.0466 s

15 2.0226 s

20 2.0066 s

25 1.956 s

30 2.002 s

hence average time period = 2.0083 s

for the second chart

Avg time = 9.444 s

hence time period = 9.444 s

for hte third chart

Length Period

0.25 4.51 s

0.5 6.92 s

0.75 8.59 s

1 9.75 s

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