M-6 HOOKE'S LAW
Bernard/Epp 4th ed experiment 16 has a good treatment of this.
If a spring is not stretched beyond its elastic limit, it obeys Hooke's law:
F = -ky
where F is the force applied to the end of the spring and y is the amount the spring stretches from its unstretched length. This is the law obeyed by a spring behaving in a perfectly elastic manner.
[Following is my rewrite of Bernard/Epp]
The brass spring is about 16 cm long and has a mass of about 170 gm. The steel spring is about 20 cm long and has a mass of 240 gm.
The limiting loading of the tapered springs is:
Steel: 1.5 kg
These values will not exceed the elastic limits of the springs.
Experiment M-4 Hooke's Law Worksheet, Fall, 1997. Brass 160 gram Hanger 50.48 gram Load Extension Period Trial Squared Trial m Squared ñ Trial T Squared m, gram x, cm T, sec T^2 x Devs. from T^2 Devs. from m Devs. 50.48 20.90 0.62 0.386 30.3 88.8337 54.241659 14.1501 0.6088244 0.0001507 90.48 24.80 0.64 0.403 34.3 90.988 58.728825 1008.14 0.7254428 0.0081799 130.48 28.80 0.83 0.694 38.4 91.2476 133.42947 8.69939 0.8257525 5.253E-05 170.48 32.60 0.92 0.842 42.4 95.374 171.39705 0.84098 0.9151323 5.142E-06 210.48 36.60 1.01 1.010 46.4 95.6399 214.67642 17.61 0.9965276 7.178E-05 250.48 40.20 1.08 1.164 50.4 103.901 254.30994 14.6684 1.0717589 5.243E-05 290.48 44.40 1.15 1.318 54.4 100.135 293.80133 11.0312 1.1420452 3.546E-05 330.48 48.00 1.21 1.469 58.4 108.584 332.68091 4.84399 1.2082496 1.483E-05 370.48 52.00 1.28 1.627 62.4 108.868 373.21336 7.47128 1.2710103 2.016E-05 410.48 56.30 1.33 1.777 66.4 102.973 411.76047 1.63961 1.3308145 4.776E-06 450.48 60.10 1.39 1.940 70.5 107.354 453.79539 10.9918 1.3880444 2.456E-05 490.48 64.10 1.44 2.074 74.5 107.636 488.0152 6.07524 1.4430063 9.038E-06 m x T T^2 (x) 1201.53 (m) 1106.16 (T) 0.0086213 STD: 3.46631 STDEV: 3.3259 STDEV: 0.0092851 Regression: extension vs mass mass vs period^2 T^2 vs m x = xo + sm y = yo + s'm Slope, s: 0.10034 1E-06 s': 257 0.1 s'': 0.00389 xo: 25.26 .001 yo: -44.9 0.1 zo: 0.1743 SSQDEV(x) 1201.53 SSQDEV: 1106.16 SSQDEV 0.0086213 k = 9766.79 dyne/cm k = 10146 dyne/cm 10148.693 m(eq) = 44.9 g m(eq) = 44.807198 Steel Load Extension Period Trial Squared Trial m Squared ñ Trial T Squared m, gram x, cm T, sec T^2 x Devs. from T^2 Devs. from m Devs. 50.48 20.20 0.33 0.106 20.1 0.01479 58.544375 65.0341 0.3141272 0.0001182 90.48 21.00 0.36 0.132 20.9 0.01916 94.666541 17.5271 0.3570713 4.262E-05 130.48 21.90 0.40 0.160 21.6 0.06513 132.44 3.8416 0.3953782 2.136E-05 170.48 22.40 0.44 0.197 22.4 0.00078 182.90782 154.451 0.4302881 0.000188 210.48 23.30 0.47 0.220 23.2 0.00789 214.54374 16.514 0.462571 4.778E-05 250.48 24.00 0.50 0.248 24.0 3.1E-05 251.37781 0.80607 0.4927432 2.275E-05 290.48 24.90 0.52 0.274 24.8 0.01498 287.52268 8.74572 0.5211716 5.705E-06 330.48 25.40 0.55 0.305 25.6 0.02586 329.8225 0.4323 0.5481276 1.9E-05 370.48 26.40 0.57 0.327 26.3 0.00314 358.7551 137.473 0.5738187 5.713E-06 410.48 27.10 0.60 0.359 27.1 0.00074 402.2889 67.0942 0.5984078 1.557E-07 450.48 27.70 0.62 0.381 27.9 0.04427 432.83265 311.429 0.6220256 2.248E-05 490.48 28.70 0.66 0.433 28.7 4.1E-05 503.20983 162.049 0.6447789 0.000172 x T T^2 (x) 0.19679 (m) 945.397 (T) 0.0006658 STD: 0.04436 STDEV: 3.07473 STDEV: 0.0025804 Regression: extension vs mass mass vs period^2 T^2 vs m x = xo + sm y = yo + s'm Slope, s: 0.01958 s': 1359 s'': 0.0007206 xo: 19.09 yo: -85 zo: 0.0623 SSQDEV(x) 0.19679 SSQDEV: 945.397 SSQDEV 0.0006658 k = 50051.1 dyne/cm k = 53651.2 dyne/cm 54785.481 m(eq) = 85 g m(eq) = 86.455731
The data above is from a student lab report, corrected because of a student blunder in computing the periods. Just to get a feel for the error dynamics, a manual fit was done by selecting slope and intercept to minimize the squares of the deviations. In the dynamic case this was done two ways (1) using the square of the period as the independent variable, and (2) using the mass as the independent variable (the prefered method, because the errors are negligible in the mass).
The static method was analyzed by writing F = mg = k(x-xo), then x = xo +(g/k)m, where the slope s=g/k, the intercept xo
The first analysis of the dynamic case used T2 = (4π2/k)(m+meq), then m = - meq + T2(k/4π2). The slope is s = k/4π2, the intercept - meq.
The second analysis of the dynamic case used T2 = (4π2/k)meq + (4π2/k)m. The slope is s = 4π2/k. The intercept is I = (4π2/k)meq, so meq = I/s.
The results for the dynamic case, by the two methods, are in good agreement, giving a spring constant for the brass spring of 10,147 dyne/cm. The mass equivalent of the spring of 44.85 gram, which is 27% of the spring's mass. The static determination of the spring constant gave k = 9800 dyne/cm, which is 2.5% smaller than the dynamic value.
For the steel spring, k÷54,000 dyne/cm. The mass equivalent is 86 g, which is 36% of its mass. The static determination the spring constant was k = 47,000 dyne/cm, 6% smaller than the dynamic value.
From Cioffari Experiment 13, Simple Harmonic Motion.
4. (a) At which point on the path of its vibration does the weight suspended on the spring have the greatest acceleration? (b) Where does it have the greatest velocity? (c) Where does it have its least acceleration? (d) Where does it have its least velocity?
[Where the word "greatest" is used, "greatest size of" is meant.]
a) Greatest acceleration at the extreme displacement positions.
8. What percent error is introduced in the calculated value of the period of bibration, for the 0.2 kg load and the 0.5 kg load respectively, if the mass of the spring is neglected?
The mass of the brass spring is 170 gm. For a 0.2 kg load,
m = 200 gm + 170/3 gm, error is 56.7/200 = 0.28, or 28%
This gets halved because of the square root, so the error is 14%. For the 0.5 kg load, the effect is 5.7%
9. If adding a certain weight in Procedure 10 would double the period, what would the mass of the added weight be?
Exp 17 in Wilson, Hooke's Law and Simple Harmonic Motion
Points were deducted from 10 as follows. Problems 1 through 5 were scored at one point each. Also one point each for each of the following which was not present, or not correct:
B5 Plot of mg vs. y for coil spring. Find k (static method)
I also looked to see if students did the following:
1. Stated and compared the two (static and dynamic) measurements of k and commented on the comparison. This must be in a separate section of the report, such as in the summary of results, or the discussion of results. One does not look for these things on the data sheets"they are results.
2. State whether the difference between the two determinations of k is reasonable in view of your estimated uncertainties for each method. Some of you thought the difference was "good" or "too large" or some other qualitative description, but this gets you no points unless you support that opinion with a quantitative discussion of the inherent errors of the experiment. Of course you can't say anything meaningful about this if you have not anticipated the need and done the necessary experimental investigations in the laboratory to investigate the sources of uncertainty. I noticed many people leaving this lab early. Clearly students had plenty of time to do such investigations, but didn't.
DISCUSSION OF ERRORS:
An error analysis is not a mere listing of possible sources of error. Some of you are too quick to attribute "bad" results to "friction effects." This is a crude cop-out to avoid thinking and analyzing the situation in detail. You can't get away with this unless you can demonstrate that friction could actually be the reason in the particular case at hand. In this experiment friction cannot affect the static determination of k at all. Friction does cause the amplitude of the "nearly periodic" motion to decrease with time, but has extremely small effect on the frequency, and therefore cannot be blamed for any poor results of this experiment. Some of you can't use your calculators properly. Question 5 could be done "in your head" with answers exactly -10, -10 and +10. Then why did I see values of 9.98, 9.86, and even 9.46? And why did I see these same dumb values on more than one paper? I know the reason: extensive plagiarism from other people's calculations. I'm seeing a lot of evidence of copying of totally wrong answers. You know that you are to do your own work, and I intend to come down hard on such plagiarism in the future. You have been warned.
1. Interpret the intercepts. Clearly the author wanted you to do more than give the values of the intercepts, or he wouldn't have used the word "interpret." Look up the word in the dictionary! It means to "explain" or "give the significance of" or "relate to something else." Even if this particular word had not been used, or even if this question hadn't appeared in the report, you should continually seek the physical significance of equations, graphs, constants, etc. as they arise.
In this case, the graph of Force vs. displacement, the intercept on the force axis (where displacement is zero) will certainly depend on what particular point you chose as "zero displacement," and what you chose as "zero force" (i.e., whether you included the weight of the hanger in the force, or not). If you followed the instructions strictly, you measured displacement from the position of the empty weight hanger, but included the hanger weight in the force. So the intercept on the force axis would then represent the weight of the hanger. The intercept on the displacement axis would indicate the displacement if the hanger were weightless (negative displacement).
2. Prove means to do a mathematical derivation of a result from basic principles. You were asked to prove that the area under the curve represents the work done. To calculate the work done is not a proof. You will find this proof done in some form in your textbook, or if not, some other textbook, so there's no excuse for anyone not doing it.
3. A surprising number of you seemed to think that the acceleration could not be greatest exactly when the velocity was zero. This demonstrates a serious misunderstanding of these important quantities. The acceleration is a maximum, and the velocity is zero, exactly at the maximum displacement. The velocity is maximum and the acceleration is zero exactly at the center point of the motion, where the displacement is zero.
4. a) y = 10 cos (2πt/T) cm
Some of you forgot that t is measured from the time you released (or pushed) the object).
5. For situation a), y = 10 cos (2πt/T)
a) y = -10 cm
6. [Not scored] Measure angles and displacements from the equilibrium position of the pendulum. For small displacements, the arc displacement is approximately equal to the horizontal displacement, and is -Rθ, where R is the length of the pendulum string. Draw a vector diagram, correctly showing the arc, and show that the tangential component of the gravitational force, mg, on the bob is mg sin θ. For small angles sin θ is approximately θ radians. So the acceleration along the arc is approximately mgθ/m = gθ. Therefore the formula for the period is:
T = 2π[-(-Rθ)/gθ]1/2 = 2π[R/g]1/2
© 1997, 2004 by Donald E. Simanek.