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 the period .637 sec and the length 10 cm, show how we find the value of a for each of the two models.

Using T = a * L^2 we plug in T = .637 sec and L = 10 cm and we get

• a = T / L^2 = .637 sec ( 10 cm)^2 = .00637 sec / cm^2.

Using T = a sqrt(L) we get

• a = T / sqrt(L) = .637 sec / sqrt(10 cm) = .201 sec / sqrt(cm).

2.  Our two models turn out to be

• T = .201 sec / sqrt(cm) * sqrt(L) or
• T = .006375 sec/cm^2 * L^2

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

• T = .201 sec / sqrt(cm) * sqrt(15 cm)

= .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

• Plugging in L = 15 cm we get

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

• What equation do we get for our observation of .637 sec at length 10 cm?

For the given information T = a * L^p gives us

• .637 = a * 10^p.

• What equation do we get for our observation of .77 sec at length 15 cm?

For the given information T = a * L^p gives us

• .77 = a * 15^p.

Write down these two equations and divide the second equation by the first. 

The equations are

• .637 = a * 10^p.

• .77 = a * 15^p.

Dividing the second equation by the first we get

• (.77 / .637) = a * 15^p / (a * 10^p)
• 1.21 = (a / a) * 15^p / 10^p
• 1.21 = 1 * (15 / 10)^p
• 1.21 = 1.5^p.

By trial and error we find that p = .475.

Substituting this back into our first equation we get

• .637 = a * 10^.475.

Solving for a we get

• a = .637 / 10^.475 = .219.

So our model T = a * L^p becomes

• T = .219 * L^.475.

• What is the resulting equation, and how many parameters does this equation have?

• Can you solve this equation for p?