dsmith@vhcc.edu

dasmith@naxs.com

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 = a * sqrt(L) gives a = T / sqrt(L) = .006375 sec / sqrt(10 cm) = .201 sec / sqrt(cm), indicating model

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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