A quantitative demonstration with springs... one that DOESN'T work!

All year long, the majority of my in-class time is spent doing quantitative demonstrations, in which the solution to an example problem is verified live in class via measurement.  Each demonstration seems to end the same way: I perform the measurement, we all see that the measurement matches the prediction within 10% or so... everyone exhales, and I say "Physics Works."

It's not a problem, per se, but by November or so the novelty of my approach has worn off.  Instead of bright shiny faces anticipating whether or not the experiment will match the prediction, I start to see resignation:  "Yeah, physics works, I know, we've done this a million times."  I've got to throw a changeup.

Once we've introduced the force and potential energy of a spring, I set up what looks like a routine quantitative demonstration.  I hang a mass from a spring, as shown in the diagram.  I determine the spring constant via the 5-second method that I discuss in this post.  This particular spring (the one on the left in the picture) has a spring constant of about 7 N/m.

Next, I hang 100 g of mass from the spring, and allow the spring to come to rest.  Using F=kx, I derive that the mass should stretch the spring by 14 cm; sure enough, a ruler placed alongside the spring shows 14 cm of stretch.  Physics works.

Now, I use the ruler to measure an additional 5 cm of stretch beyond the equilibrium position.  The question: After I release the mass from rest, what will its speed be when it passes through the equilibrium position again?

[Note to readers before you flame me about these next couple of paragraphs:  Keep reading.  Trust me.]

I use the force on a spring equation, F=kx, to calculate the force on the mass to be (7 N/m)(0.05 m) = 0.35 N.  Then, Fnet = ma, so the acceleration of the mass is (0.35 N)/(0.10 kg) = 3.5 m/s².

I take a brief interlude, if someone asks, to (correctly) explain that with a vertical spring, it's completely correct and most simple to treat it as if it were a horizontal spring, but with the x = 0 position at the place where the mass would hang at rest.  In other words, it is acceptable to consider that the spring force kx is the net force, as long as we define x = 0 at the resting position.


Now we have enough information to make a kinematics chart.  The mass starts from rest, its acceleration is 3.5 m/s² and it will move a distance of 5 cm.  Using the kinematics equation v² = v_o² + 2ax, the speed at the equilibrium position is 0.59 m/s, or 59 cm/s.

"How do we verify this prediction?" I ask the class.  They quickly and accurately tell me to place a motion detector below the mass, and to look at the velocity-time graph.  I do so, I read the v-t graph, and I see that it says about 40 cm/s.  Physics wor------  oops.  Physics DOESN'T work.


All year, I've conditioned the class to expect measurements accurate to 10 or 15%.  I'm not even getting within 30% this time... that's not a matter of "experimental error," especially when I try again and get the same result; or even more especially when I try it with the other spring, and I'm still off by more than 30%.  So, why didn't physics work?

Despite the desperate grasping at straws from some students, someone usually comes up with the right answer:  I used kinematics when acceleration was not constant.  The force of the spring was only 0.35 N immediately after release.  As the spring compressed, the spring force got smaller by F = kx.  So the acceleration got smaller.  So the mass didn't speed up as rapidly, and ended up going less than 59 cm/s.

How do we solve this problem correctly?  Using conservation of energy, of course.  Spring potential energy is converted to kinetic energy, 1/2 kx² = 1/2 mv².  Solving, I get v = 41 cm/s.  Now, physics works.

Side note:  Sure, occasionally a student will try to stop me from using kinematics with changing acceleration.  My response is a quick "shhh!!!," Dr. Evil Style, but with a smile.

GCJ

Equal length spring set

I hate doing unpaid shilling for deep-pocketed companies.  However, when PASCO provides exactly the right tool for teaching a physics concept, as they so often do, I can do nothing but spread the good word.

Take a look at the three springs you see in the picture to the right.  All are clearly equal in length and diameter.  They seem identical in every respect except color.

But when I hang a 500 g mass from each spring, the springs stretch different amounts.  The spring constants can be ranked, calculated, used for prediction of extension under a new load, etc.  What a great lab tool.

PASCO sells a five-spring set via the link here.  The red, blue, and yellow springs have a book value spring constant of 25 N/m, 30 N/m, and 35 N/m, respectively.  In my picture, the yellow spring is slightly less stretched than the blue one, but only slightly.  Well, that's okay, because (a) the yellow spring is supposed to have a larger spring constant than the blue spring, and (b) PASCO only lists the accuracy of each spring constant as +/- 5%.  It could well be that the blue spring is at the top edge of the 5% tolerance, or about 1.5 N/m too high; the yellow spring might be at the bottom edge of its tolerance, putting the stretch close to, but not quite, identical for each.

Now, for years I've used cheap-o springs that I bought in bulk from random sites online.  But these springs get stretched past their elastic limit, intertwined with other springs, and generally destroyed very quickly.  PASCO's set of five springs comes in a storage box, and these springs are tough to tangle.  Plus the color codes allow you to determine very quickly how carefully a lab group has done an experiment.  I'd suggest getting at least a demonstration set of these, if not a class set, unless your budget is lilliputian.

I always argue that physics equipment should be bought over many years, a few pieces at a time.  That's not the reality of most schools' budgeting processes, which generally give more money than you know what to do with all at once, but then nearly nothing for years on end.  That's simply not practical.  You want to be able to buy new toys when you find out about them; don't buy too much at once, spread your purchases out over the years.  

Well, my colleague Curtis bought a set of these, and now I'm jealous.  Good thing we ordered another set for me for next year...  Thanks, Curtis, and thanks, PASCO.

GCJ

What if my force vs. length graph for a spring is weird for small displacements?

Tim and Andy measuring the force applied by a spring

I think every physics class in the known universe does the F vs. x experiment for a spring:  The force on a spring is measured with a spring scale or hanging masses, and is plotted on the vertical axis of a graph.  The length of the spring (or the displacement from the resting position) is measured with a meterstick and plotted on the horizontal axis.  Because F = kx, the slope of this linear graph is the spring constant k.  

(As an aside, I've written up a detailed approach to this experiment for the College Board -- take a look here.)

This experiment is beautiful because the data are easy to take, and because even the worst experimenters get something resembling a line.  However, occasionally you'll see something weird -- the graph will be a line most of the way, but very small displacements will give a significantly steeper slope.  See the graph to the right (and click on it to enlarge if you can't quite see).  

What's going on?

First of all, quash the inevitable misconception:  "Oh, that makes sense because the more the spring stretched, the more force we had to use."  Well, of course -- that's what F = kx means.  We should need more force to stretch the spring for larger displacements.  

The slope of this graph represents the spring constant k, which indicates the stiffness of the spring. What's happening here is that the spring is significantly stiffer under about 3 cm of stretch.  Does that make any physical sense, though?

Well, in this case, yes.  If you get this sort of data, take a careful look at the spring you're using:

See how many of the coils are touching each other?  I asked the class to be very quiet... and then I began to stretch the spring a couple of centimeters.  We could all hear the "poing!" sounds of the individual coils unsticking from each other.  All the coils were fully separated when I had stretched the spring... about 3 cm.

Finding a spring constant in five seconds

In my general physics section, we’re studying springs: force of a spring, energy of a spring, and the mass-on-a-spring as an example of simple harmonic motion. Laboratory exercises with springs are useful and fun, especially because it’s easy to get good, reliable data. Before I introduced springs formally in class, I assigned a lab exercise in which they determined the spring constant of a spring. The experiment I assigned is detailed here via collegeboard.com.

Today we began a different experiment involving the spring in simple harmonic motion. Thing is, in order to use the equation

,

each lab group has to know the spring constant of their spring. Usually they know k because they use the same spring from the first experiment. But, I forgot to have them label their springs for later use. I was too caught up in the physics tournament and lab cleanup, I guess.

I asked the students to make a “quick and dirty” measurement of k just using a few weights and a measurement of how far those weights extended the spring. But I don’t yet trust my general students to do quick and accurate lab work. Estimates of k were varying by 40% within the same lab group on the same spring! What to do?

Fortunately, I had the force probe and motion detector hooked up from my in-class demonstration of the force of a spring. The setup is in the picture to the right, though I admit the picture isn't so clear. Some mass hangs from the spring, which is attached to the force probe. The motion detector sits on the table beneath the mass. The mass is allowed to vibrate, and logger pro takes force vs. distance data.

It takes only about 5 seconds to acquire a graph like the one you see in the screen shot. Press the “slope” button at the top of the screen, and you get the slope of the force vs. displacement graph – voila, the spring constant!

I’ll leave it as an exercise to the reader why the graph has a negative slope. Feel free to post a comment with the answer.

Falling behind...

It's been busy busy at Woodberry. Last week we held the US Invitational Young Physicist Tournament on campus (about which more later). Raffles Institution, from Singapore, defeated Woodberry Forest in the final.

As far as the Woodberry team was concerned, the best part of the tournament was the final evening's party at the Holiday Inn Express. Although all the teams socialized, the Woodberry guys* seemed to slather their attention heavily on the team from Brisbane Girls Grammar School. (They're GIRLS! And they have Australian accents!)

We all had great fun for the weekend. The cost of that fun, from my perspective, was a week of falling behind in my classes as I took care of details as tournament director. I have next to me this morning a stack of papers 7.5 cm high... and that's AFTER I spent two nights this week grading papers on dorm duty.
What do you do when you're so hopelessly behind that you will certainly not catch up before next week's end of the term?

Start by recognizing that you're NOT going to catch up with every assignment. In a marathon grading session, it's not worth starting at the beginning of the stack and intending to get to the end. Accept that your work will be incomplete. I picked out a few homework problems at random to grade. It's late enough in the school year that grading papers is unlikely to uncover anything new about a particular student. The diligent ones will still be diligent, the lazy ones still lazy, and the smart ones still smart. The whole purpose in grading now lies in checking up, sending the message that "I'm still watching you!" Just a few spot checks can do wonders for making sure the class keeps up with their work.

The other aspect to catching up with grading is to add as little as possible to the stack. My class starts each day with a short quiz. On Friday, I wrote a 5-question multiple choice quiz about one of the problems from the night before. Don't expect that I'll be grading that problem, now -- this quiz has evaluated their homework, and saved me considerable time.

* Woodberry is an all-boys boarding school, so in this case "guys" is not a gender-neutral term.

Here's one of the three homework problems that were assigned for Friday, which I think I got from the 1997-vintage Zitziewitz-Merrill text, but I'm not sure:

A fisherman’s scale stretches 3.9 cm when a 2.7 kg fish hangs from it.
(a) What is the spring constant of the scale?
(b) What will be the amplitude and frequency of vibration if the fish is pulled down 2.5 cm more and released so that it vibrates up and down?

And, below, take a look at the multiple choice quiz. Notice I've changed values so calculators are not necessary. (Why do the questions start at #15? Because multiple choice quizzes for the whole term go on the same scantron. That means I only have to grade multiple choice quizzes every 50 questions or so!)


A fisherman’s scale stretches 4.0 cm when a 2.0 kg fish hangs from it. The spring is pulled down 2.5 cm more and released so that it vibrates up and down.

15. What is the spring constant of the scale?
(A) 0.05 N/m
(B) 0.5 N/m
(C) 5 N/m
(D) 50 N/m
(E) 500 N/m

16. What is the amplitude of the harmonic motion?
(A) 4.0 cm
(B) 5.0 cm
(C) 6.5 cm
(D) 2.0 cm
(E) 2.5 cm

17. The period of the harmonic motion is 0.40 s. What is the frequency of the harmonic motion?
(A) 0.40 Hz
(B) 2.5 Hz
(C) 4.0 Hz
(D) 0.25 Hz
(E) 5.0 Hz

18. In a new experiment, the spring is pulled down 5.0 cm instead of 2.5 cm to begin the harmonic motion. How does the new period compare with the period in problem 3?
(A) It doubles.
(B) It remains the same.
(C) It is cut in half.
(D) It is multiplied by √2.
(E) It is divided by √2.

19. In a new experiment, a 4 kg fish is attached to the same spring and pulled down 2.5 cm to begin harmonic motion. How does the new period compare with the period in problem 3?
(A) It doubles.
(B) It remains the same.
(C) It is cut in half.
(D) It is multiplied by √2.
(E) It is divided by √2 .