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.
Using T = a * L^2 we plug in T = .637 sec and L = 10 cm and we get
Using T = a sqrt(L) we get
2. Our two models turn out to be
Which model is consistent with a period of .77 sec at a length of 15 cm?
Using T = .201 sec / sqrt(cm) * sqrt(L) and substituting L = 15 cm we get
= .201 sec / sqrt(cm) * sqrt(15) * sqrt(cm)
= .201 sec / sqrt(cm) * 3.8 * sqrt(cm)
= .76 sec.
T = .006375 sec/cm^2 * L^2
T = .006375 sec / cm^2 * (15 cm)^2
= .0064 sec / cm^2 * 225 cm^2
= 1.4 sec.
By how much does the better model deviate from our observation?
From the calculation done above it looks like the deviation is about | .77 sec - .76 sec | = .01 sec, but the actual deviation might be smaller or larger because of roundoff error.
What period does our model predict for length L = 5 cm?
Being a little careless with units we get
T = .201 sqrt( 5 ) = .201 ( 2.23 ) = .44, meaning .44 sec.
What period does our model predict for length L = 20 cm?
T = .201 sqrt(20) = ... = .88, meaning .88 sec.
What is the ratio of these predicted periods?
The ratio is .88 sec / (.44 sec) = 2.
How could we have predicted this ratio from the fact that one length is 4 times the other?
The formula is T = .201 sqrt(L).
3. If we hypothesize a 2-parameter model T = a * L^p, with parameters a and p, then
For the given information T = a * L^p gives us
For the given information T = a * L^p gives us
Write down these two equations and divide the second equation by the first.
The equations are
Dividing the second equation by the first we get
By trial and error we find that p = .475.
Substituting this back into our first equation we get
Solving for a we get
So our model T = a * L^p becomes