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course phy 201
Using your data sheet from today's class, answer the following:
What were the lengths of your two pendulums?
12 inches
17.2 inches
What were the frequencies of your pendulums?
(shorter) 53
(longer) 34
What was the ratio of the length of the longer to the length of the shorter?
17.2/12 = 1.43
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What was the ratio of the frequency of the shorter to the frequency of the longer?
53/34 = 2.21
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What is the square of the ratio of lengths, and what is the square root of the ratio of the lengths?
1.43^2 = 2.05
sqrt 1.43 = 1.12
What is the square of the ratio of the frequencies, and what is the square root of the ratio of the frequencies?
2.21^2 =4.88
sqrt 2.21 = 1.49
Find two pendulum lengths such that the frequency of one pendulum is half that of the other.
What were your lengths?
3 inches
11.6 inches
What is the ratio of your lengths?
11.6/3 = 3.87
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How did you arrive at your two lengths?
By using the frequency of two, kept adding length to the longer string (6 in, 53/37= 1.43; 7 in., 53/34 = 1.56; 9 in., 53/30 =1.7; 11.5 in., 53/27 =1.96; 11.75 in., 53/26 =2.03; 12 in., 53/26 = 2.03) used at thirty seconds. Therefore, found that the number was between eleven and twelve which is roughly around a frequency of two (that the frequency of one pendulum is half of the other pendulum).
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Well done and nicely reported.
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