1. Tabattanon 1
KamolnatTabattanon
Simple Harmonic Oscillations Lab
Introduction:
Simple harmonicoscillationsrefertothe motionof an oscillatingsystem, i.e.aspringor a light
blade,which followsHooke’slaw
F = -kx [Equation1]
where Fis the linearrestoringforce inthe system, kisthe springconstant,andx is the displacement
fromthe equilibriumline,alsoknownasamplitude. The graphof a simple harmonicoscillatinggraphis
sinusoidal.
In thisexperiment,ahacksawblade wasclampedontoawoodenblock. Massesof clay were
stuck ontothe unclampedendof the hacksaw blade. The free endwaspulledbackandallowedto
oscillate freelyforaperiodof time.
Picture 1) Lab set-up
The purpose of thislab will be tofindthe relationshipof the periodof asimple harmonic
oscillationtothe massof oscillatingsystem.
If we assume simple harmonicoscillationinthe movementof the hacksaw blade, we canapply
the formula
ω = √ (k/m) [Equation 2]

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and fromthisequation,aformulaforthe periodinrelationtothe mass can be derivedas
T2
= (4π2
/k)*m [Equation3]
where,forbothequations2and 3, T isthe periodexpressedinsecondspercycle,kisthe spring
constantof the blade in Newtonpermeter,andmis massin kilograms. Itcan be predictedthatthe
periodsquaredwill be proportional tothe mass,andthe proportionalityconstantwill be (4π2
/k).
Design:
ResearchQuestion: Whatisthe relationshipbetweenthe periodandthe massof a simple harmonic
oscillation?
Controlledfactors:
The length of the hacksaw blade allowedtooscillate waskeptconstant. At the beginningof the
experimentwhenthe bladewasclamped,the lengthwasmeasuredwithametricruler. Tokeepthe
lengthconstantthe blade waskeptinplace for the whole experiment. The particularblade itself was
paintedsothat half of the blade waspaintedred. The red half wasusedas the free end,therefore
makingiteasyto keepthe lengththe same throughout.
The heightat whichthe experimentwasperformedwaskeptconstantbykeepingthe labset-up
on a single table andnotmovingitto any otherheightlevel. The set-upwasnottiltedatanypointin
the experiment.
Althoughthe amplitudeatwhichthe blade waspulledbackwaskeptroughlythe same,the
periodof the oscillationisunaffectedbythe amplitude. Toprove this,a separate testwasdone witha
relativelyheavymassoveralongerperiodof time. The periodwascalculatedforthe beginningof the
oscillationandlateronafterdampingtookeffect. The periodwasthe same despitechange in
amplitude.

3. Tabattanon 3
Graph 1) Shows20 secondsof a heavymassat the endof the blade withdamping. The frequencywas
measuredatthe beginningandtowardsthe endof the oscillation.
Variables:
Independent:The massof the clayaddedto the endof the blade wasincreasedforeachtrial.
The mass was measuredviaelectronicbalance.
Dependent:The resultingoscillationwasgraphedwithamotiondetector. Tencycleswere
countedto calculate the periodwithreasonablyminimal uncertainties.
Procedure:
A hacksawblade wasmassedwithanelectronicbalance. Itwasthenclampedontoa wooden
blockso that half of itsfull lengthisprojectedfromthe edge of the brick,asseeninPicture 1, and this
lengthwasmeasuredwithametricruler. A motiondetectorwassetbehindthe system.
In the controlledtrial,the hacksaw blade waspulledbackandreleased. The oscillationof the
blade wasgraphed. Thisprocesswas repeatedthree times.
A piece of claywas massedviathe electronicbalance andstuckontothe endof the blade. The
ball of clay wascarefullypositionedsothatthe endof the blade reachedapproximatelythe middle of
the ball of clay. The blade wasagainpulledbackandreleased. The trial wasrepeatedthree timeswith
the same mass. A total of five differentmassesof claywere used;three trialsweredone foreachmass,
carefullypositioningthe middle of the claytoreach roughlythe endof the blade.

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Data Collectionand Processing:
Mass of Blade:14.59 ± 0.01 g
Lengthof Blade:16.0 cm ± 0.1 cm
Distance to Centerof Clayon Blade: 16.0 cm ±
0.1 cm
Mass ( ± .00001
kg)
tIn (± .02 sec) tF (± .02 sec) t10 Periods (± .02
sec)
t1 Periods (± .02
sec)
Average (±.02
sec)
0.01459 0.72 1.38 0.66 0.07 0.07
0.56 1.22 0.66 0.07
0.82 1.50 0.68 0.07
0.02256 0.55 2.10 1.55 0.16 0.15
0.50 2.00 1.50 0.15
0.42 1.90 1.48 0.15
0.03284 0.45 2.60 2.15 0.22 0.21
0.55 2.65 2.10 0.21
0.29 2.43 2.14 0.21
0.04360 0.28 2.96 2.68 0.27 0.27
0.74 3.42 2.68 0.27
0.90 3.58 2.68 0.27
0.06416 0.96 4.20 3.24 0.32 0.32
0.78 4.02 3.24 0.32
0.68 3.94 3.26 0.33
0.07730 1.14 5.20 4.06 0.41 0.41
0.82 4.88 4.06 0.41
1.08 5.12 4.04 0.40
Table 1) Raw data obtainedfromthe oscillationgraphs. Uncertaintyforaveragesis 0.01.

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Sample Graph:
Graph 2) 3rd
mass-trial 3. Frequencyobtainedbymeasuringthe time for10 cycles.
Sample Calculation:
(Time at Cycle 10- Time at Cycle 1) / 10 cycles
(3.58-0.90) ± 0.02 seconds/10 cycles= .27 ± 0.02 cycles/sec

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Graph 3) The relationshipbetweenperiodandmassof the blade
Sample Calculations:T2
Uncertainty:Mass 3, Trial 4
High:0.352
= 0.1225 => 0.12
Actual:0.332
= 0.1089 => 0.11
Low: 0.312
= 0.0961 => 0.10
Uncertainty= 0.01

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Mass ( ± .00001 kg) Period^2(± .01 sec)
0.01459 0.01
0.02256 0.02
0.03284 0.04
0.0436 0.07
0.06416 0.10
0.0773 0.17
Table 2) Mass vs the periodsquared.
Graph 4) The relationshipappearsfairlylinear.

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Graph 5) High-Lowfit. The uncertaintyforthe y-interceptcomesouttobe 0.01 and the uncertaintyfor
the slope is 0.3.
Conclusion:
The predictedrelationshipbetweenthe massof the blade and the periodof a cycle was
proportional,wherethe proportionalityconstantwas(4π2
/k). Inthe final graph,however,the
relationshipwasindeedlinear,butitwasnot proportional. The reasonforisthat the mass of the blade
was neverzero. Inthe derived[Equation3] the periodcouldhave onlybeenzeroif the massof the
blade waszero. The blade cannotachieve zeromass,andtherefore,the graphmustexhibitay-
intercept. The acquiredequationforthe relationshipis
T2
= (2.4 ± 0.3)*m – (0.03 ± 0.01) [Equation4]
The slope inthe final graphcan be usedto findthe k,springconstant,of the hacksaw blade
used. The original massof the blade withoutanyclayaddedshouldbe usedalongwith[Equation3]

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Sample Calclation:k
T2
= (4π2
/k)*m
.072
= (4π2
/k)*0.01459
k = 117.549
Uncertaintyfork
High:(.07 – 0.02)2
= (4π2
/k)*(0.01459 - .00001)
k = 230.238
Actual:.072
= (4π2
/k)*0.01459
k = 117.549
Low: (.07 + 0.02)2
= (4π2
/k)*(0.01459 + .00001)
k = 71.159
The uncertaintyis80. The large uncertaintymakesthe kvalue very unreliable.
The y-interceptonthe graphrepresents the periodwhenthe massof the blade iszero. The
mass of the blade can, inreality,approachzero,butit can neverbe trulyzero.
The x-interceptrepresentsthe effective massof the blade (the half thatisallowedtooscillate).
Whenthe periodiszero,the blade isessentiallynotoscillating.
The equationderived inthisexperimentaswell asthe proportionalityof the periodsquaredand
the mass can be appliedtoobjectsinthe earth’satmosphere andfollowsHooke’sLaw.
Evaluation:
Weaknessesof thislabinclude the shakingof the woodenblockthatthe blade wasclampedon,
the uncertaintyof the centerof mass of the clay,and the instrumental errorof the motiondetector.
The shakingof the woodenblockwouldhave includedthe blockintothe systembeingstudied.
The movementwould nothave beensimple harmonicoscillation. Tofix thisproblem, the woodenblack
can be properlyclampeddown,oradifferentmeansof clampingthe blade downcanbe used. For
example,asturdyextensiontoatable wouldbe heavyenoughnottomove withthe blade.
The centerof masseffectshowquicklythe blade oscillates. The furtherbackthe clayispushed
alongthe blade, the clearerthe difference becomes. The clayusedwasshapedintoa ball. Perhapsa

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more accurate shape to have usedmayhave beena square,where the centerof masscouldbe
measuredmore accurately. The centerof massin thisexperimentwasapproximatedtobe the endof
the blade. It wasimpossibletosee where the centerof massactuallywas. A see throughmass like a
clearball witha sloton the side or a clear gel matterwouldmake iteasiertosee the center.
The motiondetector’ssample rate maxedoutat50 samplespersecond. Evenso,the
uncertaintyforthe labtrialswas lessthanthat of the instrumentitself. A betterqualitymotiondetector
can minimize the uncertaintyevenmore.