Intermediate Laboratory PHY 266
Popular in Course
Miss Lucienne Hamill
verified elite notetaker
verified elite notetaker
GEOL 1200 - 003
verified elite notetaker
verified elite notetaker
verified elite notetaker
verified elite notetaker
Popular in Physics 2
Miss Lucienne Hamill
verified elite notetaker
This 24 page Class Notes was uploaded by Miss Lucienne Hamill on Wednesday October 28, 2015. The Class Notes belongs to PHY 266 at Wake Forest University taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/230740/phy-266-wake-forest-university in Physics 2 at Wake Forest University.
Reviews for Intermediate Laboratory
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/28/15
PHY 266 Projectile Motion and Numerical Integration 1 Trajectory with no air resistance A A canon ball is fired straight up with an initial velocity v0 Assuming no air resistance write an expression for the position of the ball as a function of time yt B Now open the Excel le plotting the trajectory Look at the different columns and see how they are calculated For one pair of columns y and V are calculated analytically using the equation from part a For the next pair y and V are calculated numerically This is done as follows First and initial position y0 and velocity v0 are defined Knowing the net force on the ball the acceleration may be calculated using Newton s 2nd law a0 FO m After a short interval of time dt we can estimate the new velocity by assuming the acceleration was constant over that interval vl v0 a0 dt Likewise we can nd the position at this time ya W t V0 dt This process is then repeated over and over again to find later values of a v and y an Fn m vn1 vn an dt yn1 yn vn dt Currently the data only runs to 16 seconds To extend the calculation just copy and past the last row This is numerical integration Speci cally this particular method is called a Left Riemann Sum The process is illustrated below graphically Note that as the interval dt decreases the accuracy of the estimate improves dt Area Av 0 at Area Ay V t II Canon ball trajectory with air resistance Now that we have a basic scheme for doing numerical integrals it is straightforward to do other problems which are difficult or even impossible to solve analytically Modify this problem to include air resistance The force of drag is given by a 1 A F pv2ACd v p density of air or 1292 kgm3 v velocity A cross sectional area of the projectile Cd drag coefficient 01 for a smooth sphere A Assume the mass ofthe ball is 10 kg the initial velocity is 200 ms and the radius of the ball is 01 m B On the same graph plot the trajectory of the ball with and without air resistance What is the maximum altitude for each case C Increase the initial velocity to 400 ms What are the new maximum altitudes III Rocket trajectory with air resistance Now consider a rocket with an initial mass of 5 kg red from rest straight up The rocket engine burns for 3 seconds and consumes 5 kg of fuel ie the rocket loses 5 kg of mass over 3 seconds The thrust of the rocket while it is burning is 100 N Assume the drag coefficient is 025 and the radius of the rocket is 01 m A Plot the trajectory of the rocket with and without air resistance What are the maximum altitudes B Increase the thrust of the rocket to 400 N Plot the new trajectories of the rocket What are the new maximum altitudes PRECESSION of a GYROSCOPE L l 11 PURPOSE to measure the precession rate ofa g roscope and compare the measured mic to theoretical expectations Tl EEORY A gyroscope is a rigid system of particles symmetrical about an His and rotating about that axis The symmetry about the axis guarantees that components of the angular momentum Lt of particle k which are perpendicular to the axis of rotation will he cumpensulcd by an equal and opposite component due to the diametrically opposite particle Thus the angular momentum L for the entire body points along the axis ofrutationi To compute 1 consider the angular momentum dL arising front dm located a distance l from the axis dL r x dp dmr x v But v is petpendicular to r so v or and dL drn r2 u If one integrates dL and dm over a disc ofradius R and mass M one obtains L MRZmZ 10 u l where l MRzZ is the moment of inertia of the disc In analogy to Newton s First and Second Laws L remains constant unless acted on by an external torque T dLdt r 2 Assume a gyroscope is balanced in the horizontal position 0 90 in Figure l 1 below In the Figure the z axis is vertical and the angle 45 gives the direction ol L in the xy plane lt rl L is a constant z 1 I lmh mquot av 9 1 0 thyFA quotC i t d l mg side view top View Flgure 11 Torque Applled to Horlzontal Gyroscope lfa small mass m is attached to the end olilhe gyroscope shaft at a distance I from the point of support the balance is upset and 10 The torquc exerted by mg about the point ofsupport is given by r r x F 3 III mgd direction oft is perpendicular to L and in the xy plane As a result I changes the direction oil but not its magnitudet This causes L to rotate about ll 2 is The expected rate precession ofL can be computed For small changes in the direction ofL Lo 11 Thus the rate ofprecession Q ddadt is given by 7 mga 4 Q mgd mgtl 5 L In If 0 90 i r mgdsin 0 instead of mgcl W A Strobe light See How a stroboscope worksquot taped to the benchtopt Familiarize yourself ilh the use ofthe strobe light to measure at fora rotating object Note that the strobe light has four ranges High vs Low and Direct vs Slow Use the motor of known speedt3500 rpm ft to check the calibration ofthc strobe at f1 W2 W3 W4 and WE It should be within 5 Be sure that you know how to convert the strobe readings from rpm revolutionsmin to radianss the standard units ofn in physics B Calculate the moment of inertia for the large disk C Level the gyroscope base as follows gyroscope axle rotated 90 gytoscope axle QDDg relatth counterweight disk adjust this iool rst then adjust this loot rotating I i disk eveltng leet Figure 2 Leveling the Base Leveling the Base D l urposely make the apparatus unbalanced by mov Rotate the gyroscope 90 degrees so the gyroscope mg the 900g counterweight towards the center axle is parallel to one side of the A and adjust the other leveling foot until the shaft will stay in this 2 Adjust the leveling foot on one of the legs of the l posttton See Figure 2 base until the gyroscope disk is aligned over the leveling foot on the other leg of the base See Fig 9 Adjust the position of the 900g counterweight until ure 2 the gyroscope is balanced without the a on mass The 30g counterweight can be used to ne tune the balance D Balance the gyroscope about its point of support by moving the cylindrical counterweights along the shah Experiments mgd I both a and Qmusl be determined ll shard to measure both at the a To test Eq 5 Q same time because u is not a constant friction slows down the spinning disk One way to make the task easier is to separate the determination of a from that ol39Q 1 With 8 90 and no added mass m start the balanced gyroscope spinnng h39 pulling him on the string Measure 1 every 105 for 605 Repeat Several times Determine u vs 2 for an average pull on the string 2 Start the balanch disk spinning in the some way as for Part l but this time add mass m In front end ot the gyroscope Measure the precession rate Q Repeat several times Note the direction ofprcccssion Le clackwisc or counterclockwise looking from the top down ls the direction as you expected S the magnitude eluse tn expcctutiuns Hint be sure tn measure the leveramt d 3 Repeat Pan 2 but place In on the back end ofthc gyroscope Note direction of precession carefully Are your observations in accord with expectations Be sure to remeasure cl 4 Install the second large disk on the gyroscope Set the two disks rotating in opposite directions Add mass m to the front of the gyroscope Observe Explain Writeup explain your measurements and qualitative observations in parts 4 File l66gyt osc0pe701doc HOW a stroboscope WOTkS Stroboscopic effects are produced when an intense repetitively pulsed light illuminates a rotating object If light ashes at frequency f are directed upon a rotating or vibrating object oscillating at frequency t the object will appear stationary iff f This apparent cessation of motion called the stroboseope effect occurs because the object is illuminated at precisely the same phase ol each cycle This apparent cessation ofmotion will also occur when the light frequency is a submultiple of the movement frequency ft 2 F lS etc 1Play around with the stroboscope Observe what happens with a known steady rotation such as a 3500 rpm motor shaft with red and yellow quotflagsquot What do you observe when n where n234 7000 rpm etc What do you observe when fis a submultiple of f1 that is l l 1n such as 1750 rpm Do you understand what you observe It is easy to make a mistake when determining the apparent frequency of a rotating object with a stroboscope 2 How to measure an unknown freguency it l lt fmax Set the stroboscope to u then slowly decrease f until the motion frcezesquot The highest f which quotfreezesquot the motion is t 3 How to measure an unknown frequency fl ifyou re not sure 139 I lt fnm for your stroboscope You have to proceed more carefully in this case because the highest available frequency which freezes the motion may be a submultiple of f fin where n is unknown A Start at the highest available strobe frequency measure the highest possible freeze frequency Call it f B Slowly decrease f until the motion again freezes Call the resultant frequency fn i C Calculate fl from the relation 1 11 mint n fiat This relation can be derived by noting that fquot n r fln 1 and solving the two equations for ft The Electronic Brazing Company strobe light has 4 ranges A switch on lhc back can in set lo quotdirectquot or quotslowquot mode providing 1 00 X at 10 X multiplier for the dial reading The knob on the side mms lhe strobe on but also permits selling lhe strobe 10 High or Low dial range The resultant ranges n rpm are Low High Slowx10 60 370 300 500 Direclx1 600 3700 3000 15000 Note that in physics problems we always work in llerlzcyclcss or radianss rather than revolutionsminuta File fcourscwork l 65 l 66 strob67200 doc RIGID BODY ROTATIONAL MOTION TORQUE ANGULAR MOMEN l UM MOMENT OF INERTIA The purpose ofthis lab is to test eoneepts oftnrque moment of inertia and angular momentum as applied to a rigid body using the Pasco Rotational Dynamics apparatus This apparatus has an accurate readout and lawfriction air bearings Most of your measurements should agree with expectations to better than 500 The rotation ot a rigid body about a xed ais is most conveniently described in terms of angular position 0 angular velocity u and angular acceleration 1 c 15 r in radians l Gs 1 I dSll tn rads 2 x a aimd in rads 3 If the torque on the body is constant a is constant and m 14 m 4 t V at Y uni llut39 5 m39 ml l ZuEl 6 0 mu 139 ML2 7 I I v CO For a pntnt P located a distance r trom the axis the magnitude nt39the tangential linear velocity is v or 8 v P t It IS convenient to write the linear acceleration a of point P intn radial and tangential components at at 0a a 1 o r 9h The rotational kinetic energy nt a rigid body is given by KM mine2 10 where I is the rotational inertia nfthe body de ned by IIr3dmirzpdv It 2 ln rotational motion torque 1 plays the role which F does in linear motion I is a measure of the turnng action ofa force F on a rigid body For rotation about a xed axis and for forces with no components pttrallel to that LKiSi r is given by er1Fsium 12 The angular acceleration ol39a rigid body about rt Fixed axis is given by 21 la 13 in direct analogy to Newton39s Second Law A ANGULAR ACCELERATION UNDER CONSTANT TORQUE Experiment 1 Seep 56 ofthe Pasco manual Note that the disks t closely on the stationary center shaft forming an air bearing for low friction Be gentle when placing disks on the center shaft so as not to damage the surfaces First place the steel disk with 58quot center hole onto the airbearing spindle then place the large aluminum disk on top ot39the steel disk note there are 3 disks use the right ones Measure the angular acceleration Cl for the Al disk with the smaller pulleyr 127 cm and a mass m of about 1525 g including 5 g for the hanger hanging from the end of the string Compute l for the large Al disk b39 sed on its mass M and its inner and outer radii r and r2 from I tr1393quot 2 14 The t Sign in 14 might seem wrong but isn39t see appendix 1 M is given in it Table tit the end of this writettpt You need to measure the radii and compute I Cmnputc the torque r from the tension 1 in the string and the leverarm r This isn t simply mgr because the tension isn39t mg Consideration of the acceleration ofm leads to the following expression for T T 2 mg 3115 I 1 J17 I where r is the radius of the pulley through which the torque is applied The tension in the string is NOT equal to mg because m is accelerating This equation is derived on p 6 of the Paseo manual Check that the measured value ot a agrees with the value predicted by Eq 13 The results should agree within i5 Experiment 2 Repeat experiment 1 with m 40 60 and 80g Test whether a is a linear function ofT Should a he a linear function ofnt39 Experiment 3 Repeat experiment lbut the larger pulley with r 254 cm and m 20g Does the pulley radius matter Why B CONSERVATION ofANGULAR MOMENTUM The angular momentum L for a nonrelativistic particle is given by Lrgtltp1mrxv l9 If r and v are perpendicular rigid body motion xed axis this reduces to L miv 30 Dynamics enters into angular motion in analogy to Newton39s second law F dpdt for linear motion through the relation 1 dLdt 2 I In this relation r and L must be de ned with respect to the same axis For a rigid body of mass M constrained to rotate about a xed axis the total angular momentum is obtained by breaking M down into little mass elements dm and integrating dL over the whole body dL rdm v 22 Because v uri L ml r39 din In 23 where l is the rotational inenia introduced in eq 1 l 4 From eq 2 i I Z 0 L is a constant rmd angular momentum is emisen39ed This is true also for item ofinteracting particles or rigid bodies as long as there IS no external torque We will test th in experiment I Experiment 4aSec Expt 4 on p 5 of the Pasco manual for details on setting the two valve pins and air pressure it s tricky With the large stccl disk on top and able to rotate freely with respect to the lower steel disk set the top disk spinning at 24 revolutionss while keeping the lower disk stationary Measure m his de nes the initial angular momentum Ll While recording to pull the valve pin From the top disk so that the top disk quotfallsquot onto the bottom disk creating a new combined rigid body with I l hump Measure tor the angular velocity ofth combined body Compute the new angular momentum LI from lgand compare it to L4 Has angular momentum been conserved Has kinetic energy been conserved Recall that Km 39Itnz alumni4m Experiment Repeat 4a with the aanninum disk instead ofthe second steel on top 4 ltl 3 C Veri cation that d rzdm SE n t is asserted in Eq 1 I that the contribution ofa mass element lm to l is proportional to 7 7 The next experiment is designed to test this Iquot dependence Expertment Using the apparatus with two sliding masses af xed to a transverse rod measure the contribution of the sliding masses to L as described in Experiment 6 of the Paseo manual Test whether dI is proportional to r or r Suguested procedure For a given position r oftlte small masses m the total moment ofinertia I is given by l I U Zmrnt where I is the moment of inertia ofthe assemth without the two masses 7 tu In Zmr39 4 Measure 1 for r loll v z quot224 r 5 values ofr Plot rot against 139 and against r Does either plot give a straight line Which one and why Minor point to consider I changes whe change Pretty tricky See Eqn 5 n I changes even though the mass in the end ofthe string doesn39t Dam needed in several experiments Object mass Inner Outer Torque g radiuscm radiuscm radiuscm Steel disklmvcr 1338 Steel diskupper 1352 Aluminum disk 463 small pulley l0 large pulley 36 steel cylinders 204 experiment 5 Appendix I Moment tifilierlia Ufa disk ofmass M outer radius r untl hole radius r Lei h be lhc thickness oftlie disk and p its density Think ol tlie disk as 1 series of rings ufrurlius r and mass rim L1 r dm rquot pdlrquot Bu 11 litmlr so I l hp II Jd Irhprquot 1quot 2 But M Znhp Irdr Jill70 rl 5011 Z 739 rl Ierf 3 A l 5 Using your computer to take data in Physics 166 lab GENERAL SETUP 1 Hardware connections a Computer off Pasco interface off b Install SCSI card in bottom slot of computer c Connect Pasco box to SCSI card using heavy at black cable 1 Turn on Pasco interface e Wait 10 s then turn on computer f Plug the transducers for the experiment into Pasco box 2 Start DataStudio a Click on New Experiment b Tell the software which transducer you are using Motion detector for SHO Rotational dmamics for torque lab Force detector for rocket c Double click 011 sensor icon to set the sampling rate required for all d Set up graphs for realtime readings 3 Acquire data By now you should know how to click on Start Stop etc 4 Save data a save DataStudio le b export data as txt le for use by Excel 5 End experiment 3 Exit from DataStudio program b Exit from Windows c Turn off your computer 1 Turn offPasco interface e Remove SCSI card from your computer File Pascolnterfacedoc Measurement ofG by Cavendish39s Method The gravitational constant G is rather difficult to measure with precision in the laboratory The trouble is that the gravitational forces between masses of laboratory size are extremel small and thus a very delicate apparatus is needed to detect these forces Henry Cavendish 73 l 1810 an English experimental physicist and chemist used a torsion balance similar to that used earlier by Coulomb to determine what we know as Coulomb s law We will use a torsion balance also since it gives good results for G and also illustrates some principles of rotational mechanics Because the apparatus is so delicate it is important that you understan t e experimental procedure so that you can carry out the measurements correctly the rst time If you e a mistake you will need to come in some other time because it takes several hours for the apparatus to settle down A diagram of the torsion balance is shown below Two equal large masses M are brought near the two small masses m attached to a rigid beam suspended by a thin ber The gravitational To U IE w39 Light beam t Large Masseslt r4 Positton I 15 kllogram Lgtge Masses Posnion I E 4 nth lm quotc Figure 1 The Gravitational Torsion Balance h 1 It r H r T manm I r5 L lllrLMiJ v gt attraction between each small mass and the neighboring large mass located 1 distance b away is given y Fr G MIDbl I This force tends to rotate the beam connecting the two small masses about the ber axis The beam rotates until the torque generated by Fr is exactly balanced by n restoring torque generatcd by the twisted ber lf Fugt M m and b can be measured G can be determined The hard part is to determine F This is done by measuring the de ections ofthc beam when the large masses M are switched from position I to position ll When you come into the lab the system will be in equilibrium with masses M in position 1 the ber is twisted clockwise and the laser betuu is at spot So on the scales The suspension ber with beam and small masses m form a torsional pendulum lfynu switch the masses M to position 1 the beam will oscillate with damped harmonic motion until it reaches the new equilibrium in which the ber is twisted counterclockwise as shown schematically below The laser beam will then be at new position Sam the 503 e Figure 3 Graph at Small MassOscillations V u L cx T ll l at Q is mutemarl uth l wu l L ILA39 L r rf laden hem Tina EC Duil5wum IN li39i l rlki EC Lt itbmum E3 Lla Theory With the large masses in Position 1Figttre4 the gravita tional attraction F between each small mass m and its neighboring large mass M is given by the law of universal avitntion FGm Mitt 11 This force exerts t torque t on the system n 1quotquot 751 12 Since the system is in equ ihrium the twisted torsion hand must be supplying an equal and opposite torque This torque Thad is equal to the torsion constant for the band K times the angle through which it is twisted e or T 46 my 13 Combining equau39orts 11 12 and 13 and taking into account that IFquot 4 gives K6 ZdGm Mbquot Rear ranging this equation gives an expression for G Bb G K de 1 1 1 4 To determine the values of B and K d l only unknowns in equation 14 it is necessary to observe the oscillations of oscillate until it nally slows down and comes to rest at a new equilibrium position At the new equilibrium position the torsion wire will still be twisted through an angle 9 but in the opposite direction of its twist in Position 1 so the total change in angle is equal to 26 Taking into account that tlte angle is also doubled upon re ection from the mirror see Figure 6 46 ASL at 9 ASML 15 in The tetsion constant can be determined by obsening the period T of the oscillations and hm using the equation quotlquot 4161Mquot 16 whoe l is the moment of inenia of the small mass SySlEm The moment of inero39a for the mirror and support system for the small masses is negligibly small compared to that of the masses thmnselvm so the total inem a can he normed as I 2m clquot L7 Therefore Kan m allT LB Substimnng equations 15 and 18 into equation 14 gives 1911Asz 19 0 T I All the variables on the right side ofequation 19 are known or measurable d 50 mm b H5 mm m15g 2150 2 L a Measure as in Figute4 By measuring the total de ection of the Light spot AS and the period of oscillation 139 the Value of G can therefore be determined Procedure 1 To begin the masses should be arranged in Position 1 Figure 2 Lb balance should be leveled and zeroed and the small masses should be at equilibrium 2 Tum on the light source and observe Lhe zero point of the balance for several minutes Io be sure the system is at equilibrium Record the zero point 5quot as accurately as possible and indicate any variaLion over time as part of your margin of error in the measurement the case which would disturb the system 4 Immediately after rotating the swivel supporL observe the light spot Record the position of the light 5th S and the time 1 every 15 secondsfor 39 about 45 minutes or until the oscillations have Stopped Analysis quotlav ledQ l Construct a graph of light spot postion versus time with time on the horizontal axis as in Figure 3 i imrml 7 From your data measure AS the change in position of it 39 quot h quot quotvi to its nal equilibrium position Efr 3 From your graph measure the period LT ofthe oscillations of the small mass system F r best results e 39 the average value ofT over Several oscilla donsi in 39 MJI utE I lrhx A tygt39 539 4 Use your results and equation L9 to determine the value of G St The value calculated in slzp 4 is subject to the follow 39 ere is amacted not Figure 7 the vector arrows shown are not proportional to the acmal forces Figure 7 Correciing the Measure Value 0 G The force F 39 is given by the gravitational law which translates In tlus case to Gm F b3 4cm and has a componentfthat is opposite to the direction of the force F 61711me b1 4d b1 441 quot2 This equation de nes a dimensionless parameter 3 that is equal to the 12lle of Lhc magnitude or 1 to that of F0 Using the equaan Fo Gmlnyb it can be deten39rtined that B tilbl 4dgt39 From Figure 7 FaFBfFu3FOFO1 4393 where F is the value of Lhe force acting on each small sphere from both large masszs and Fu is the force of attracuon to Lhe nearest large mass only Similarly in L10 0 61 13 where o is your experimentally deter mined value for the gravitational constant and GD is mmrmt systema Finally on Gl gt13 Use this equation mm equation 110 to adjust your measured value ANALYSIS using EXCEL SOLVER Using Excel s Solver function do a least squares t to the entire data setexeluding the data points for KO thereby datemiining AS and T more accurately Here39s how Under the assumption that the angular displacement ofthe pendulum from its equilibrium position is small the restoring torque provided by the ber is proportional to the angular displacement TK3 13 Em 39C I dzeldtz 50 153P 111 This is the equation ofmntion ofa simple hanunnic oscillator The solution is BAcoswl 112 A e 113 m I In practice there is damping ofthe harmonic motion so that 639Aexp ylcosmt 114 where Use Excel to fit your raw data to Eq 114 and obtain leastsquares values ofA y a and 45 See Using Excel to t a damped sinusoid to experimental dataquot Compare the value ofA to the value of AS you estimated manually Use the value ol39m to rccotnpute the period T ofthe oscillator and compare your new value to the value ofT estimated manually Remember that a erf 27rT Recompute G using the new values In your writeup include a plot of y tted and ynbserved against 1 If you have worked carefully the two curves will be in excellent agreemenL showing that this torsional pendulum behaves exactly as predicted by the equation of motion for an underdampcd harmonic oscillator File GOl doc in Cuserdatewordcourseworkkl65166 39Rea IS nil 305 V 35 H IN gt an tmezj Ht FKVILL PL 9 MM mm LL Cltill LhL1lLL LI 352 5iun H a 3352 5146 556 U w 7 39 2 quot A 33 7 quot39 927J3 7 z lt 39S 7 SLS VS rsgr 5 iii 3 Z t m m C LJU MPH Using Excel to fit a damped sinusoid to experimental data Short version 39lhc gnztl is to lintl thc parametch in a model cquzttinn which best lits n sct ulcxpcrintental tlill pointsl The model equation ie y A coswr Eq I l l lacu your cxperimcntul observations in two culumnst and y0 s 1 attache From Scicnchnrkslmn use out and paste mark dutztl 2 quotrents additional columns VII in the Excel spreadsheet NTL C CNTL Vt Ll nquot and quoterrorquot to hold yvalucs computed by Eq I and deviations between equation tlnd cxpcrinmnl 3 Set up names or the target cells to be adjusted in well as the key parameter I In twu short columns Vuritthle Name Varitthlu Value A C turgct cell or A gamma S C tttrgct cell for gamma w 3 C turch cell for w phi etc nL chi squared The unknnuns A gamma 1 mid phi cnrrespund in nlwintts mt In the parameters nl39cq l 39l he utlditionnl phascirtclur phi and baseline Bl are needed because liqv 1 requires that y A when 1 0 and t when t to Next link 11 target cell to each Vttriablc Nttmc ltt hnltl the 5 unknowns as I39nllows To Create it target ccll fnr variable A Click on the latrgct cell 7 Nztmc 3 Dclinc Deline tts quotA del39tult s rt Fallow thc same prucechtre to create target cells l39ur gamma wt phi tnd BL l l39inlcr appmximme values in the target cells created in step 3 For example Al etc Leave the Cl tliSllll dl Ctl box empty 5 l39pe in the procedure by hich l Veel should compute u lit in Cnlumn C ll39the first cell uh lit is c nd the rim cell on i A2 39 39 Iick on 2 Enter the formula In compute 3 lit for the model Eq I translated into Ewell A EXPtgammnquotA2c05w kA2phi BL Press enter nr click on the green check in the tnnllmr he value olyilit should upp t ll39 in 2 Cup the thrmuln in CB intu the remaining cells in the yi t cnlumn 39 a t39omnute the errnrquotcolumn D by de ning D2 by the l39omiula 2 Bl then copy the tbnunlu to the remaining cells in the Error culunm D 7 Compute y 39l39llIS S T lL KEY l39ARAMIi39l39ER Wl llCH Wll Bl MINIMIZED T0 GE l39llF quotlll S ll 1quot 21 m r l39 39 easil computed by applying FVcel s UMSQ function it the quoterrorquot column already created Deline the target cell for chigsquaretl as 81 lMSQtlZlNN Nnte DNN is the lust cell in column 1 K Now HSh erel39s SOLVlCR routine to minimize click nn the turget cell for Chi squared Tools 2 Solver 1 check that lnrgel cell fur x1 is entered Lug i7 I click on quotminimizequot In minimize 7 c enter the eunrdinnles nflhe target eells for A gummu V phi illttl BL for example G2JG0 INotc ll needed 1 click on quotsolvequot NC values ofmrget cells A gnmmu w phi BL and 1 and new ulues ol39y lit should uppeur ll they do nut Us to help trace the logic ul39the euleulatinns Nole k3 needed y should be computed in n lew seconds c the quotAuditinlf function under quot l39nnlsquot 0 Plot Lulu and Llit l I It Sample spreadsheet altnelted File ExcelSnlvergtampedllldnc Spreadsheet C lllISLird IRCXCClPll 105lblt isllh erxls y t 4 635 3607 2209 0788 0310 0827 0648 0170 1415 2778 3925 4582 4600 3988 2905 1625 0461 0 307 0 500 0083 0831 2010 3164 4012 4356 4121 3377 2 314 1195 0294 0175 70108 0469 1406 2469 3396 3967 4049 3633 2 829 1 840 0910 error 0003 70063 0100 0085 Variable name A gamma w phi BL chiisq uared variable value 3 002431702 0200349885 4000010462 0499691809 2000107403 0488118324 Solver Fit to Damped Sinusoi O yobs yjit Luuoms d l0 0 o o o U 139 o 9 o 39 fquot o o o o o N o 39 o o 39 o o 39 o 39 o o 0 517777797 0 times
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'