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Hooke’s Law and Simple Harmonic Motion

↘︎ Feb 22, 2010 … 4′ … download⇠ | skip ⇢

Purpose

To determine the spring constant of a spring by measuring its stretch versus applied force, to determine the spring constant of a spring by measuring the period of oscillation for different masses, and also to investigate the dependence of period of oscillation on the value of mass and amplitude of motion.

Hypothesis

If the applied force (mass) is to remain the same while the vertical displacement is increased, there period will remain the same. However, if the vertical displacement is held constant while the applied force in increased, the period will increase. This is in accordance with the derived equation T = 2π (M / k)0.5.

Labeled Diagrams

See attached sheet.

Data

Part I

M (kg)

Mg (N)

y (m)

0.0249 0.244 0.0550
0.0449 0.440 0.101
0.0649 0.636 0.157
0.0849 0.832 0.212
0.0949 0.930 0.251
0.105 1.028 0.270

k = 3.53 N/m

Part II

A (m) ∆t1 (s) ∆t2 (s) ∆t3 (s) ∆t avg. (s) αt (s) T (s)
0.0200 7.17 7.38 7.40 7.32 0.127 0.0736
0.0400 7.40 7.42 7.33 7.38 0.0473 0.0273
0.0600 7.33 7.38 7.44 7.38 0.0551 0.0318
0.0800 7.33 7.31 7.29 7.31 0.0200 0.0115
0.100 7.10 7.14 7.08 7.11 0.0306 0.0176

Part III

M (kg) ∆t1 (s) ∆t2 (s) ∆t3 (s) ∆t avg. (s) αt (s) T (s) T2 (s2)
0.0500 7.55 7.63 7.55 7.58 0.0462 0.0267 0.758
0.0600 8.10 8.15 8.13 8.13 0.0252 0.0145 0.813
0.0700 8.75 8.85 8.73 8.78 0.0643 0.0371 0.878
0.0800 9.37 9.34 9.36 9.36 0.0153 0.0088 0.936
0.0900 9.95 9.94 10.00 9.96 0.0321 0.0186 0.996
0.100 10.50 10.48 10.46 10.48 0.0200 0.0115 1.05

ms = 0.00940 kg

Slope = 10.5 s2/kg

Y-Intercept = 0.0687 N

k = 3.75 N/m

C = 0.694

% difference = 6.04%

Graphs

Part I

Part II

Part III

Questions

1. Do the data from Part 1 verify Hooke’s Law? State clearly the evidence for your answer.

The data correlate close to Hooke’s Law, but not quite. The law states that F = -ky, where F is in this case Mg and y equals the negative displacement. After graphing forces versus displacement, a value of 3.53 N/m was determined as the spring constant. However, when applying this value to the equation and using recorded displacement values, the calculated force come up less than the actual for used. For example, in the first trial y = -0.055 m. Multiplying that value by the extrapolated spring constant gives a theoretical force of 0.194 N, but the actual force used was 0.244 N. All other trials yield a similar lowball theoretical force.

2. How is the period T expected to depend upon the amplitude A? Do your data confirm this expectation?

Period is not expected to depend upon amplitude, as suggested by the equation T = 2π (M / k)0.5, where amplitude is absent as a variable. The data confirms this expectation, as the period was nearly the same for each trial.

3. Consider the value you obtained for C. If you were to express C as a whole number fraction, which of the following would best fit your data (1/2, 1/3, 1/4, 1/5)?

The obtained value of C is 0.694, which is closest to 1/3.

4. Calculate T predicted by the equation T = 2π (M / k)0.5 for M = 0.050 kg. Calculate T predicted by T = 2π ( [M + Cms] / k )0.5 for M = 0.050 kg and your value of C. What is the percent difference between them? Repeat for a value for M of 1.000 kg. Is there a difference in the percent differences? If so, which is greater and why?

T predicted using the first equation and M = 0.050 kg is 0.726 s. T predicted using the second equation and M = 0.050 kg is 0.771 s. This is a percent difference of 6.01%.

T predicted using the first equation and M = 1.000 kg is 3.24 s. T predicted using the second equation and M = 1.000 kg is 3.26 s. This is a percent difference of only 0.615%. The greater percent difference occurs at the lower weight because the weight of the spring is almost insignificant at higher weight. The proportion of the mass to the spring is so great that it has almost no effect on the calculation.

Conclusion

During part one of the experiment, the vertical displacement of a spring was measured as a function of force applied to it. The starting position of the spring was recorded using a stretch indicator. Mass was added to the spring, and the displacement was recorded. This was repeated with various amounts of mass. From these data, a graph of force versus displacement was plotted, and a linear fit slope revealed the spring constant. In this endeavor, the spring constant was valued at 3.53 N/m.

However, when applying this spring constant to the recorded displacements in the Hooke’s Law equation, the calculated forces are lower than the recorded forces. In similar manner, when rearranging Hooke’s Law to solve for displacement, the calculated displacements are larger than the actual recorded displacements. This means there was human error, most likely in terms of not being precise with the displacement readings because the recordings for the masses used were accurate. Because such small masses were used, any error in displacement readings was augmented. The spring used may also not have been perfect.

During part two of the experiment, the period of the spring was measured as amplitude changed while mass remained constant. The period remained nearly the same throughout every trial, which was to be expected. Any differences in period may be accounted to inadequate stopwatch usage and inaccurate starting displacements throughout the trials. It should be noted that at amplitude of 0.100 m, the hook lost contact with the spring for a split second at the apex of oscillation, which accounts for its oddity in period. This was something that could not be avoided; at that amplitude the spring pulled the hook up too quickly which caused the loss of contact. The resulted graph of period versus amplitude yielded a linear fit slope of close to 0 (-0.247 s/m), which was predicted.

During part three of the experiment, the period of the spring was measured as mass was varied while amplitude remained constant. As the mass was increased, the period also increased. This was not surprising considering the given equations. The square of the period versus mass for each trial was plotted and a linear fit was taken. The slope and y-intercept of this line was then used to determine the spring constant and C, the fraction of the spring’s mass that should be taken into account for the equation T = 2π ( [M + Cms] / k )0.5. The calculated value of k was 3.75 N/m, which is only 6.04% different from the value determined earlier of 3.53 N/m. The value of C was determined to be 0.694, which is closest to the whole number fraction of 1/3. Any error during parts two and three can be attributed to inaccurate stopwatch recordings and slight variance in displacement and release of the masses at each amplitude.

Equations

T = 2π (M / k)0.5

T = 2π ( [M + Cms] / k )0.5

slope = 4π2 / k

y-intercept = 4π2Cms / k

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  • 10 Feb 22: Hooke’s Law and Simple Harmonic Motion #Dr. Paul J. Angiolillo #PHY 1042 (General Physics Lab II) #Saint Joseph's University
  • 10 Feb 11: Tinkering with Tin #CHM 2521 (Inorganic Chemistry Lab) #Dr. Peter M. Graham #Saint Joseph's University
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  • 10 Feb 1: Strokes of Genius #ambition #psychology #sport

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Tinkering with Tin: Synthesis of SnCl(CH2C6H5)3 and SnCl4[OS(CH3)2]2

↘︎ Feb 11, 2010 … 6′ … download⇠ | skip ⇢

Tinkering with Tin: Synthesis of SnCl(CH2C6H5)3 and SnCl4[OS(CH3)2]2

Abstract

The reaction of tin with benzyl chloride under reflux yields SnCl(CH2C6H5)3 in low percent yield (around 15%) due to various factors. 1H NMR of SnCl(CH2C6H5)3 shows a singlet at δ 3.15 with satellites containing coupling constant between Sn-H of 78.1 Hz. The reaction of SnCl4 with OS(CH3)2 yields SnCl4[OS(CH3)2]2 with poor percent yield (around 140%) due to various factors. The IR spectrum of SnCl4[OS(CH3)2]2 shows a large S-O stretch at 899.7 cm-1, which is a lower frequency than the S-O stretch of 1042.5 cm-1 from the IR spectrum of OS(CH3)2, suggesting coordination on the metal complex at oxygen.

Introduction

The reaction of tin with benzyl chloride yields the metal complex SnCl(CH2C6H5)3. The reaction specifically takes place in the following manner:

[Figure missing.]

This is compound of interest because of the different isotopes of tin that naturally occur. In order to determine the structure of said substance from a 1H NMR spectrum, the peaks must be analyzed for satellites occurring from coupling of hydrogen to isotopes of tin. The observed spectrum is a combination of SnCl(CH2C6H5)3 molecules containing different isotopes of Sn, namely 117Sn and 119Sn, among others. The value for JSn-H between satellites can confirm the presence of SnCl(CH2C6H5)3. Mass spectrum can also be used to confirm the presence of SnCl(CH2C6H5)3 in the product by looking for a peak at m/z equal to the molecular mass of the substance (427 g/mol).

SnCl4[OS(CH3)2]2 can be synthesized from the following reaction:

[Figure missing.]

It is a reaction of interest because it can be used to determine where OS(CH3)2 coordinates to Sn. Coordination of S to Sn would result in a higher frequency of S-O stretch on an IR spectrum than the frequency of the S-O stretch on the IR spectrum of OS(CH3)2 due to a strengthening of the S-O bond. Coordination of O to Sn would result in a lower frequency of S-O stretch because of a weakened S-O bond from the resonance form needed to secure that coordination.1

Experimental

All syntheses were carried out in air and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of SnCl(CH2C6H5)3 (1) and SnCl4[OS(CH3)2]2 (2) were based on reports published previously.1

SnCl(CH2C6H5)3 (1). 325 mash-Aldrich Sn powder (1.939 g, 16.34 mmol), 99% Aldrich benzyl chloride (6.0 mL, 52 mmol), and deionized H2O (4.0 mL) were added subsequently to a 50 mL round-bottom flask. A small stir bar was added then added to this solution. A sand bath was placed over a stir plate and a cold water condenser with greased joint was inserted into the round-bottom flask containing the solution. This connection was further secured with a keck clip. The round bottom flask was placed in the sand bath and the condenser was connected to the cold water source.

Once secure, the sand bath was set to 50% power and the stir bar was spun a shade over moderate speed. The solution was allowed to reflux for 2.75 h. Once reflux was complete, the 50 mL round-bottom flask was removed from the condenser and allowed to cool in an ice bath until it was cold. The liquid was decanted from the white precipitate and discarded. The white precipitate was saved. Ethyl acetate (10 mL) was added to the precipitate and this solution was heated using a sand bath at 65% power and stirred at moderate speed until the white solid had completely dissolved. The round-bottom flask was then taken off the heat and submerged in an ice bath for approximately 0.5 h to allow for recrystallization.

The stir bar was removed and a high vacuum was used for about 10 minutes to extract the extraneous liquid and leave a white powder. Several mL of ether were added to the round-bottom flask and a glass stir rod was used to break up the chunks of powder and dissolve it in the ether. This solution was then suction filtered with a 30 mL glass frit and allowed the dry. The precipitate collected was 1 (0.471 g, 13.49% based on the amount of Sn powder used). 1H NMR (CDCl3): δ 7.25 (s, CDCl3), 7.19 (s, Ph-H), 7.17 (s, Ph-H), 7.05 (s, Ph-H), 3.15 (s, JSn-H = 78.1 Hz, CH2).

SnCl4[OS(CH3)2]2 (2). SnCl4 (2.25 mL, 19 mmol), anhydrous diethyl ether (45 mL), DMSO (2.9 mL, 41 mmol), and ether (5 mL) were subsequently added to a 125 mL Erlenmeyer flask. The DMSO was added using a syringe for safety purposes while the other reagents were measured and added using graduated cylinders. This solution yielded a precipitate which was isolated by suction filtering the solution through a 50 mL glass frit. Several extra mL of ether were added to the Erlenmeyer flask to help aid in transfer of all the precipitate to the filter. Once dry, the white powder precipitate 2 was collected and weighed. 11.04 g (26.49 mmol) were recovered, giving a yield of 139.4%. FTIR (ATR) ν(S-O) 899.7 cm-1 (s S-O coordination to Sn).

OS(CH3)2 (3). The IR spectrum of (3) was taken by Dr. Graham. FTIR (ATR) ν(S-O) 1042.5 cm-1 (s, S-O linkage).

Results

The reaction of Sn, benzyl chloride, and H2O yielded 0.471 g of the product, SnCl(CH2C6H5)3. This translated to 1.102 mmol, and thus was a 13.49% yield based on the amount of Sn used, which was the limiting reagent in the reaction. Sn reacted to form the product in a 2:1 ratio, and 16.34 mmol of Sn was used to start, so that proportion was taken into account when calculating the percent yield. Proton NMR yielded several peaks. A set of peaks were found at δ 7.19, 7.17, and 7.05 were indicative of phenyl resonances derived from protons on the phenyl ring of the product.1 A sharp peak located at δ 7.25 was due to the chloroform solvent 1. Coupling between Sn and H produced a singlet at δ 3.15 with satellites 1JSn-H equal to 78.1 Hz due to the presence of isotopes 117Sn and 119Sn.1 The mass spectrum of SnCl(CH2C6H5)3 showed a peak at m/z = 427.0,1 which is the molar mass of said substance.

The reaction of SnCl4 and DMSO yielded 11.04 g of product, SnCl4[OS(CH3)2]2. This translated to 26.49 mmol, and thus was a 139.4% yield. IR spectrum of SnCl4[OS(CH3)2]2 showed its largest peak at 899.7 cm-1 while the IR spectrum of OS(CH3)2 gave its largest peak at 1042.51 cm-1.

Discussion

The percent yields for each product are less than stellar. During the synthesis of SnCl(CH2C6H5)3, letting the reagents reflux for a longer time period, closer to 3 hours, may have been beneficiary to resulting in more product. The reagents may not have all completely reacted. Some of the precipitate may have been accidentally removed during decanting after reflux, and much ethyl acetate may have been added to that precipitate. A minimal amount should have been used while trying to dissolve the white solid precipitate in the sand bath. The less ethyl acetate needed and used would have resulted in a better percent yield.

The fact that recrystallization did not seem take place as detailed1 and that a vacuum had to be used to dry the product most likely did not help the yield of product either. Product may have been lost during the vacuuming process. The glass frit used for filtering was not of the utmost quality, and so product may have escaped during that process, also. The product was also not completely dry, and gave a misleading mass measurement, meaning the percent yield is even lower than recorded. The product obtained did seem to give a clear 1H NMR spectra, meaning it was fairly pure. The coupling between Sn and H due to isotopes 117Sn and 119Sn 1 in the product is distinctly visible around δ 3.15 with a JSn-H of 78.1 Hz. The NMR chemical shifts of the Sn-CH2 protons and the C­6H5 protons of SnCl(CH2C6H5)3 are so different because of the 117 and 119 isotopes of Sn. They affect the coupling with H, producing satellites. The C6H5 protons are not affected by this coupling. The literature states that peaks for phenyl resonance appear around δ 7 and chloroform appears around δ 7.24, which seems to validate the experimental values obtained (δ 7.05 to 7.19 for phenyl resonances ad δ 7.25 for chloroform).1

The percent yield of SnCl4[OS(CH3)2]2 is likely thrown off because the product was not dry when it was weighed. Several extra mL of ether were used for transport of the precipitate to the glass frit for filtering, which would most likely lead to loss of product. Not all of the precipitate was able to be transferred from the Erlenmeyer flask to the frit. Product was distinctly lost due to a seemingly defective frit, too. Attempts to refilter the filtrate were unsuccessful. The fact that the frit quality was poor meant that the product was not able to be dried well and thus carried extra weight. The product seemed to smoke away as it was allowed to dry further in exposure to air, and lost mass over time. This conundrum could not be adequately explained. IR spectrum of the product SnCl4[OS(CH3)2]2 shows its largest stretch at 899.7 cm-1, hinting that the coordination to Sn occurs at oxygen, as this frequency is lower than that of the largest stretch from the OS(CH3)2 spectrum (1042.51 cm-1). This is due to resonance and weakening of the S-O bond.1 The two spectra are similar save for the shifting of that one peak. The stretch at 1042.51 cm-1 is consistent with the value stated in the literature for S-O stretching (approximately 1100 cm-1).1

Conclusion

The main purposes of the experiments were to decipher the 1H NMR for SnCl(CH2C6H5)3 in order to determine JSn-H and to interpret the IR spectra of OS(CH3)2 and SnCl4[OS(CH3)2]2 to tell whether O or S coordinates to Sn. The JSn-H was found to be 78.1 Hz. The observed spectrum is a combination of SnCl(CH2C6H5)3 molecules containing different isotopes of Sn, namely 117Sn and 119Sn, among others. This is what makes the 1H NMR difficult to interpret, but the JSn-H value is validation of the presence of SnCl(CH2C6H5)3. The percent yield for that reaction was poor, due to varying factors. Longer reflux time, less ethyl acetate used during recrystallization, and better filtration techniques all would have contributed to an improved percent yield.

The IR spectrum of OS(CH3)2 showed its largest stretch around 1050 cm-1 while the IR spectrum of SnCl4[OS(CH3)2]2 gave its largest stretch around 900 cm-1. This lower frequency of stretch suggests that O and not S coordinates to the Sn. This is due in part to the resonance form needed to form the complex, which weakens the S-O bonding, and thus lowers the stretching frequency. The yield for the reaction to form SnCl4[OS(CH3)2]2 was also poor, which can be mostly attributed to poor filtering and drying techniques, along with the overuse of ether.

References

(1) “Synthesis and Techniques in Inorganic Chemistry,” Third Edition, G. S. Girolami, T.R. Rauchfuss, and R. Angelici, University Science Books, 1999.

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Jeans Progress…

↘︎ Feb 3, 2010 … 1′⇠ | skip ⇢

I have 1 month left to go in my year long jean contest. Might not be the winning pair, but I think they look alright:

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Strokes of Genius

↘︎ Feb 1, 2010 … 2′⇠ | skip ⇢

Roger Federer won the Australian Open over the weekend, his 16th Grand Slam event, putting him now two ahead of Pete Sampras for the most career major titles by a mens tennis player.

What may be even more impressive than Federer’s number of victories is his streak of reaching at least the semifinals or better in majors…22. That adds up to be over 5 years where no one could beat him until the final 4 of the tournament. That kind of consistency is unheard of…even great players like Sampras and Borg never achieved anything close to that.

The old heads will tell you that Rod Laver was better than Fed, but there is no denying his greatness. I feel so fortunate that I have been able to watch his career unfold.

Some may tell you it’s his world class forehand that makes him so good, or maybe his ability to ever-so effortlessly glide around the court, but I think there is no doubt that his success is in large part thanks to what sits between his ears.

Federer has been mentally stronger than any other player in tennis the last half dozen or so years. There is just no other way to explain his success.

Earlier in his career, Fed may have been more physically gifted than some of his peers, but that is no longer the case. Federer is getting older while his opponents are getting younger. They can hit harder, run faster, and play longer than him…in terms of raw ability, they can do pretty much everything better than him on the court.

However, these players lack one aspect to their game that Federer dominates…and that is the mental aspect of tennis. His ability to focus on the court has allowed him to overcome anything his opponents may have on him physically.

Anyone that has played tennis knows how difficult it is to maintain your composure on the court…you can hit 10 great shots in a row, miss an easy one, and totally lose focus because of that one bad miss-hit. You need to have the ability to treat each point almost as a new match…you can’t worry about what happened 30 seconds ago.

You will seldom see Federer get rattled…I could probably count the number of times I’ve seen him get flustered over the years on only one hand. Nothing seems to bother him when he’s on the court, yet other players will noticeable become frustrated or lose focus in the key moments of a match.

Federer also has the ability to rise to the occasion during big points, whereas other good players are not able to (at least as consistently). I can’t even imagine what it’s like playing in front of a packed stadium like they do, but I am sure every action within the match is magnified when thousands of people are oooing and ahhing. It must be tough!

How has Federer raised himself to this level of mental prowess? I wish I could answer that myself. His opponents are the best in the world, and I am sure are nearly as mentally tough as him, but for whatever reason, they have not quite matched his ability.

I do know that he puts a lot of work in off the court, conditioning himself for long matches (even though he often wins in straight sets). I suppose that he comes into every match knowing that he is in the best shape he could possibly be in, but I am sure nearly all of his opponents could say they are just as prepared.

It would be interesting to know what Federer is thinking while he plays and how he fortifies his mental game…maybe he’ll write a book and explain how he does it. Until then, all we can know for certain is that he is a prime case and point for mind over matter.

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ADAM CAP is an elastic waistband enthusiast, hammock admirer, and rare dingus collector hailing from Berwyn, Pennsylvania.

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