http://vhmthphy.vhcc.edu/
T = a * L^2
T = a * L^.5
| Length (cm) | cycles | Time Interval (sec) | period (sec / cycle) | frequency (cycles / sec) |
| 10 | 124 | 78 | 0.629032258 | 1.58974359 |
| 10 | 123 | 78 | 0.634146341 | 1.576923077 |
| 10 | 121 | 78 | 0.644628099 | 1.551282051 |
| 10 | 125 | 78 | 0.624 | 1.602564103 |
| 10 | 125 | 78 | 0.624 | 1.602564103 |
| 10 | 124 | 78 | 0.629032258 | 1.58974359 |
| 10 | 122 | 78 | 0.639344262 | 1.564102564 |
| 10 | 116 | 78 | 0.672413793 | 1.487179487 |
| 10 | 120 | 78 | 0.65 | 1.538461538 |
| 10 | 124 | 78 | 0.629032258 | 1.58974359 |
| 10 | 122 | 78 | 0.639344262 | 1.564102564 |
| 10 | 123 | 78 | 0.634146341 | 1.576923077 |
| mean period 0.6374 sec | ||||
| 15 | 102 | 79.5 | 0.779411765 | 1.283018868 |
| 15 | 103 | 79.5 | 0.77184466 | 1.295597484 |
| 15 | 104 | 79.5 | 0.764423077 | 1.308176101 |
| 15 | 102 | 79.5 | 0.779411765 | 1.283018868 |
| 15 | 103 | 79.5 | 0.77184466 | 1.295597484 |
| 15 | 102 | 79.5 | 0.779411765 | 1.283018868 |
| 15 | 102 | 79.5 | 0.779411765 | 1.283018868 |
| 15 | 103 | 79.5 | 0.77184466 | 1.295597484 |
| 15 | 105 | 79.5 | 0.757142857 | 1.320754717 |
| 15 | 103 | 79.5 | 0.77184466 | 1.295597484 |
| 15 | 102 | 79.5 | 0.779411765 | 1.283018868 |
| mean period 0.7733 sec |
period is seconds / count. For example 78 second / (124 counts) = .63 seconds / count.
frequency is counts / second. For example 124 counts / (78 sec) = 1.6 counts / second.
We obtain the following results from our observations:
We consider two possible mathematical models for pendulum period vs. length. One model is T = a * L^2, the other is T = a * sqrt(L), where T is the period in seconds and L is the length in cm.
For each model an appropriate value of a is to be determined from our data. Our data relate the variables T and L. The quantity a in each model is a parameter to be determined from the data.
Each of these models is a one-parameter model. We wish to find which one-parameter model works better for our pendulums.
There are various ways to determine the most appropriate value of a for each model. Here we start by finding the value of a which corresponds to our data for the 10-cm length, where the period was .6375 second:
T = a * L^2 gives a = T / L^2 = .006375 sec /(10 cm)^2 = .006375 sec / cm^2, indicating model
- T = .006375 sec/cm^2 * L^2.
T = a * sqrt(L) gives a = T / sqrt(L) = .006375 sec / sqrt(10 cm) = .201 sec / sqrt(cm), indicating model
- T = .201 sec / sqrt(cm) * sqrt(L).
We now check each model with the data from the 15-cm pendulum:
T = .006375 sec/cm^2 * L^2 gives us T = .006375 sec / cm^2 * (15 cm)^2 = 1.44 sec, approx.. This does not agree well with our observation of .77 sec.
T = .201 sec / sqrt(cm) * sqrt(L) gives us T = .201 sec / sqrt(cm) * sqrt(15 cm) = .78 sec, approx.. This agrees well with our observation of .77 sec for the 15-cm pendulum.
Quiz for 0825
1. In class last time our observations indicated that a pendulum of length 10 cm has a period of .637 sec and a pendulum of length 15 cm has period .77 sec.
We hypothesized that one of the two models T = a * L^2 or T = a sqrt(L) would give us a good prediction of the period T of a pendulum vs. its length L, with period in seconds when L is in cm.
2. Our two models turn out to be
Which model is consistent with a period of .77 sec at a length of 15 cm?
By how much does the model deviate from our observation?
What period does our model predict for length L = 5 cm?
What period does our model predict for length L = 20 cm?
What is the ratio of these predicted periods?
How could we have predicted this ratio from the fact that one length is 4 times the other?
3. If we hypothesize a 2-parameter model T = a * L^p, with parameters a and p, then
Write down these two equations and divide the second equation by the first.
... get 5 cm observation and use the three to get a quadratic model, then see which works better for a 20 cm pendulum.
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