Pendulums
Practice
practice problem 1
 What is the length of a seconds pendulum at a place where gravity equals the standard value of 9.80665 m/s^{2}?
 What is the period of this same pendulum if it is moved to a location near the equator where gravity equals 9.78 m/s^{2}? How much time does the pendulum lose or gain every 30 days?
 What is the period of this same pendulum if it is moved to a location near the north pole where gravity equals 9.83 m/s^{2}? How much time does the pendulum lose or gain every 30 days?
solution
The period of a simple pendulum is described by this equation.
T = 2π √ ℓ g Make length the subject.
ℓ = gT^{2} 4π^{2} Put numbers in.
ℓ = (9.80665 m/s^{2})(2 s)^{2} 4π^{2} Get answer out. (Keep every digit your calculator gives you. This is a test of precision.)
ℓ = 0.993621386 m
Note how close this is to one meter. In the late Seventeenth Century, the the length of a seconds pendulum was proposed as a potential unit definition. Since gravity varies with location, however, this standard could only be set by building a pendulum at a location where gravity was exactly equal to the standard value — something that is effectively impossible.
Let's calculate the number of seconds in 30 days. This part of the question doesn't require it, but we'll need it as a reference for the next two parts.
t = 30×24×60×60 = 2,592,000 s
Back to the original equation. Length and gravity are given. Period is the goal.
T = 2π √ ℓ g Weaker equatorial gravity in.
T = 2π √ 0.993621386 m 9.78 m/s^{2} Slightly longer period out.
T = 2.002723096 s
Half of this is what determines the amount of time lost when this pendulum is used as a time keeping device in its new location.
T_{½} = 1.001361548 s
Dividing this time into the number of seconds in 30 days gives us the number of seconds counted by our pendulum in its new location.
n = 2,592,000 s 1.001361548 s n = 2,588,476 periods
That's a loss of 3524 s every 30 days — nearly an hour (58:44).
Repeat.
T = 2π √ ℓ g Stronger polar gravity in.
T = 2π √ 0.993621386 m 9.83 m/s^{2} Slightly shorter period out.
T = 1.997623207 s
Divide by 2.
T_{½} = 0.9988116033 s
Divide this into the number of seconds in 30 days.
n = 2,592,000 s 0.9988116033 s n = 2,595,084 periods
That's a gain of 3084 s every 30 days — also close to an hour (51:24). Pendulum clocks really need to be designed for a location. Support your local horologist.
practice problem 2
A classroom full of students performed a simple pendulum experiment. They attached a metal cube to a length of string and let it swing freely from a horizontal clamp. They recorded the length and the period for pendulums with ten convenient lengths. Use these results to determine the acceleration due to gravity at this location.
solution
There are two basic approaches to solving this problem graphically — a curve fit or a linear fit. Let's do them in that order.
First method: Start with the equation for the period of a simple pendulum.
T = 2π √  ℓ 
g 
Compare it to the equation for a generic power curve.
y = Ax^{n}
If we let…
y = T  and  x = ℓ 
then…
A = 2π/√g  and  n = ½ 
Set up a graph of period vs. length and fit the data to a square root curve.
Use the constant of proportionality to get the acceleration due to gravity.
g = 


g = 


Second method: Square the equation for the period of a simple pendulum.
T^{2} = 4π^{2}  ℓ 
g 
Compare it to the equation for a straight line.
y = mx + b
If we let…
y = T^{2}  and  x = ℓ 
then…
m = 4π^{2}/g  and  b = 0 
Set up a graph of period squared vs. length and fit the data to a straight line.
Now use the slope to get the acceleration due to gravity.
g = 


g = 


Bonus solutions: Start with the equation for the period of a simple pendulum.
T = 2π √  ℓ 
g 
Solve it for the acceleration due to gravity.
g =  4π^{2}ℓ 
T^{2} 
Compute g repeatedly, then compute some basic onevariable statistics. This method isn't graphical, but I'm going to display the results on a graph just to be consistent.
The most popular choice for the measure of central tendency is probably the mean (g bar).
g = 9.678 m/s^{2}
But the median is also appropriate for this problem (g tilde).
g̃ = 9.722 m/s^{2}
How about some rhetorical questions to finish things off?
9.742 m/s^{2}, 9.865 m/s^{2},
Which answer is the right answer? All of the methods used were appropriate to the problem and all of the calculations done were error free, so all of them. Which answer is the best answer? That's a question that's best left to a professional statistician. What is the answer supposed to be? The answers we just computed are what they are supposed to be. The problem said to use the numbers given and determine g. We did that. What is the generally accepted value for gravity where the students conducted their experiment? I think it's 9.802 m/s^{2}, but that's not what the problem is about.
practice problem 3
solution
Answer it.
practice problem 4
solution
Answer it.