PHYS 210 Lab Reports Solved, Fall 2006

Date: 03/01/2007 Physics 210L Interference & Diffraction Name: Section number: 3 Partner’s Name: Instructor: a- Diffraction : Distance from slit to screen: D = 0.772 m. Slit width (mm) Separation Between two Dark Fringes (mm) Wavelength of laser  ( nm ) 0.04 12.7 658 12.9 668 0.08 6.71 695 6.63 687 0.16 3.52 730 3.41 707 Estimate the average value of  (show a sample calculation) and its rms error. Comment. ±nm WE HAVE A SLIGHT ERROR HERE, WHICH ISN'T GREAT COMPARED TO THE AVERAGE VALUE OF THE WAVELENGTH. THIS ERROR COULD COME FROM HUMAN ERROR, WE PROBABLY DIDN'T MEASURE THE DISTANCE X ACCURATELY. THE MACHINES COULD HAVE ERROR, THE CONDITION OF THE ROOM COULD AFFECT THE RAYS OF THE LASER. BUT ALL IN ALL THE RESULTS ARE PRECISE AND THE AVERAGE VALUE IS ACCURATE AND IT ONLY HAS A 1.62 % ERROR WITH THE ORIGINAL VALUE. Describe qualitatively what would you see if you use a white light instead of a laser. A WHITE LIGHT IS NOT MONOCHROMATIC. SO THERE WOULDN'T BE THE SAME PHASE DIFFERENCES FOR EACH RAY OF LIGHT COMING OUT OF THE SLIT. BECAUSE OF THIS YOU WILL NOT SEE THE PATTERN OF WHITE LIGHT AND DARK LIGHT AS WE SAW WHEN WE HAD THE LASER AS A LIGHT SOURCE. 1 Grade: b- Interference: Distance from slit to screen: D = 0.768 m. Double slits X (mm)  ( nm ) w = 0.04 mm d = 0.25 mm 2.20 716 2.10 684 2.00 651 2.23 726 2.31 752 w = 0.04 mm d = 0.50 mm 1.10 716 1.20 781 1.30 846 1.31 853 1.29 840 w = 0.08 mm d = 0.25 mm 2.22 723 2.10 684 2.21 719 2.19 713 2.17 706 Estimate the wavelength of the laser and its rms error. Comment. ±nm THE SOURCES OF ERROR HERE ARE THE SAME AS THE PREVIOUS EXPERIMENT. THE MACHINES COULD HAVE SOME MINOR DEFICIENCIES, THE SLITS MAY NOT BE EXACTLY 0.04 OR 0.08 MM APART. WE MOST LIKELY MADE SOME ERRORS WHILE CALCULATING THE DISTANCE X. ONCE AGAIN THE ROOM MAY BE A BIT HUMID AFFECTING THE RAYS. OUR CALCULATED AVERAGE WAVELENGTH VALUE HAS AN 8 % ERROR WITH THE THEORETICAL VALUE. ALSO, THE VALUES THAT WE CALCULATED ARE PRECISE. Explain how does the slit width affect the value of X? FROM THE PREVIOUS TABLE WE SEE THAT X IS NOT AFFECTED WHEN THE WIDTH OF THE SLIT IS INCREASED OR DECREASED. HOWEVER THE NUMBER OF FRINGES DECREASES ON THE SCREEN IF AS THE SLIT WIDTH CHANGES. NOW THE REASON WHY X DOESN’T CHANGE AS THE SLIT WIDTH CHANGES IS BECAUSE THERE IS A CONSTANT RATIO FOR FINDING X, SO AS ONE FACTOR CHANGES ANOTHER FACTOR MUST CHANGE TO BALANCE OUT THE RATIO. By what percentage, would the separation between two adjacent fringes change if the experiment is performed using an Ar+-laser emitting at 514 nm? IF WE USE THE CONDITIONS WHERE W = 0.04 MM AND D = 0.25MM THEN THE VALUE X FOR THE ORIGINAL LASER AT A WAVELENGTH OF 680 NM, IS 2.08896 MM. WHILE THE 2 VALUE OF X FOR THE ARGON CATION LASER IS 1.579008 MM. THE PERCENTAGE BY HOW MUCH THE SEPARATION BETWEEN TWO ADJACENT FRINGES WOULD CHANGE IS 24.4 %. 3 PHYS 210 L The Spectrometer OBJECT: …. TO MEASURE THE WAVELENGTH OF LIGHT EMITTED BY A MERCURY AND WHITE LIGHT SOURCE. THEN WE MEASURED THE INDEX OF REFRACTION FOR DIFFERENT WAVELENGTHS (COLORS ON THE SPECTRUM). GRATING: I. Incandescent lamp: Illuminate the slit a) Describe the spectrum after looking at it. THE SPECTRUM HAS CONTINUOUS COLORS AND EXACTLY LIKE THE SPECTRUM IN THE LITERATURE. WE SAY THE COLORS IN THIS ORDER: VIOLET, BLUE, TURQUOISE, GREEN, YELLOW, ORANGE, RED, PINK, BLACK. b) Angles measured in the first order: Extreme red: 24.1 degree Extreme violet: 14.9 degree c) Wavelengths:  Extreme red: 680 nm Extreme violet: 428 nm II. Mercury Lamp: a) Describe the spectrum you observe and conclude: YOU SEE IN THE SPECTRUM QUANTIZED COLORS AS OPPOSED TO A CONTINUOUS SPECTRUM IN THE PREVIOUS EXPERIMENT. THE COLORS ARE IN STRIPS. THERE ORDER IS AS THEY COME IN THE FOLLOWING TABLE. b) Sharp line I II III IV V VI Angle (deg) 14.5 15.5 17.5 19.5 20.7 22.6 Color Light Purple Blue Light Green Green Yellow Red Wavelength 417 nm 445 nm 501 nm 556 nm 589 nm 640 nm Prism: Mercury spectrum Number of orders per color: 3 The most deviated color: Violet Minimum Deviation: Color () Violet (417 nm) Blue (445 nm) Green 556 nm) Yellow (589 nm) Orange (615 nm) Angle 40.9 40.5 40 39.7 39.5 n() 1.542 1.537 1.532 1.529 1.526 Plot n() versus . n(Lambda) Vs. Lambda 1.525 1.526 1.527 1.528 1.529 1.53 1.531 1.532 1.533 1.534 1.535 1.536 1.537 1.538 1.539 1.54 1.541 1.542 1.543 0 100 200 300 400 500 600 700 Lambda n(Lambda) Date: 25/10/2006 Physics 210L Atwood’s machine A) Constant total mass. Trial M1 (Kg) M2 (Kg) aexp (m/s2) Fnet (N) M1 +M2 (Kg) atheory (m/s2) % Diff. Run #1 0.0450 0.0550 0.880 0.0981 0.100 0.981 10.3 Run #2 0.0400 0.0600 1.82 0.196 0.100 1.96 7.14 Run #3 0.0350 0.0650 2.73 0.294 0.100 2.94 7.14 Run #4 0.0300 0.0700 3.71 0.392 0.100 3.92 5.36 Run #5 0.0250 0.0750 4.62 0.491 0.100 4.91 5.91 Plot aexp versus the net force? Using a linear regression deduce the slope of your graph and the error on it? What is the slope of this curve, Comment? If we plot this graph with the Net Force against the experimental acceleration, the slope of this graph will be the total mass of the system. (mass=force/acceleration) However due to experimental errors the value won’t be exact. Slope = 0.105 kg = 105 g Error = 5% Our result is accurate but the reasons for this error are: Human error, friction between the pulley and string, the screws of the pulley may be rusty. There are others but these are the obvious ones. For example temperature may play a role in this error. B) Motion down an inclined plane. First incline: Sin = 0.0747  = 4.28 degrees. Trial Distance (meter) x Time (seconds) t Acceleration (m/s2) a = Run # 1 0.500 1.23 0.661 Run # 2 0.500 1.24 0.650 Run # 3 0.500 1.24 0.650 aAverage = 0.654 ± 0.004 Second incline: Sin = 0.115  = 6.62 degrees. Trial Distance (meter) x Time (seconds) t Acceleration (m/s2) a = Run # 1 0.500 0.598 2.80 Run # 2 0.500 0.598 2.80 Run # 3 0.500 0.599 2.79 aAverage = 2.797 ± 0.003 Third incline: Sin = 0.0813  = 4.67 degrees. Trial Distance (meter) x Time (seconds) t Acceleration (m/s2) a = Run # 1 0.500 1.07 0.873 Run # 2 0.500 1.08 0.857 Run # 3 0.500 1.07 0.873 aAverage = 0.868 ± 0.005 Plot the acceleration (a) versus sin. Using linear regression deduce the value of the slope? What is the slope of this curve? Compare with the theory. Slope = 54.5 m/s2 Error = 456 % The slope that we calculated came out to be 54.5 m/s2 which should be the gravity 9.81 m/s2. However the value is very far off from the real value which is 9.81 m/s2. The reason for this is because the time that we calculated for the second incline was incorrect. When we were calculating the time for all the other inclines, we placed the first sensor at 60 cm and we released the glider from the 50 cm line. This was done for all of inclines, however for the second incline we did something different. Since this incline was very high the first sensor couldn’t fit at the 60 cm line so we moved it to the 70 cm line, and we dropped the glider from the 50 cm line. This in turn created a big problem which we did not realize until now. Since the glider had a bigger distance to travel before passing under the sensor means that it was able to gain a lot more speed after rest, compared to the other inclines. Therefore we know where our mistake is. Now error can be caused by many things. In this case there may still be friction between the glider and air track, which would produce a wrong value for the acceleration. Another error is human error, and this error is explained, in detail, in the preceding paragraph. Another source of error could be that the sensors used aren’t that precise, so this would create a technical error. PHYSICS 210L Exp. VII Conservation of Energy Object: In these experiments we were conducting various experiments to test the theory of the conservation of energy. In the first experiment, we saw how the work done by the weights was all stored in the rubber band as elastic potential energy. In the second experiment we saw how elastic potential energy stored in the rubber band, released all this energy into the glider as kinetic energy, hence the glider gained velocity. In the third experiment we were testing to find out what the force of friction was on the track, and were able to see that it was approximately constant. Finally in the fourth experiment, we saw how inelastic collision happened. If it were to be elastic collision then all the energy should have been transported from glider one to glider two. However in this case, not all the kinetic energy was conserved. A-1. Elastic potential energy of a rubber band: Weights N 1.962 N 1.472 N 0.981 N 0.491 N 0.1962 N 0.0981 N Displacement of glider in m 0.025 m 0.019 m 0.013 m 0.007m 0.0025m 0.002 m Plot F (x) against x Work done to stretch the rubber band: 0.0210 J Elastic potential energy of the stretched band: 0.0210 J A-2. Work-Energy: Mass of glider 0.18325 kg Length of glider 0.129 m Trial I II III IV reading of timer 0.308 s 0.305 s 0.305 s 0.309 s speed of glider 0.419 m/s 0.423 m/s 0.423 m/s 0.417 m/s Kinetic energy of glider: 0.0162 J This result is close to the answer in part A, and the percentage difference is 30.0%, which is alright. A-3. Friction on the Track: Location of glider I II III IV speed of glider in m/s 0.402 m/s 0.397 m/s 0.391 m/s 0.381 m/s Grade: Estimate the force if friction: 0.00125 N Comment: The force of friction between each distance is constant in the first two while it is approximately equal in the last distance, compared to the others. Therefore we have been able to record accurate readings. The velocities are decreasing gradually because of the force of friction, and if there were no friction then the glider MUST have the same velocity at every instant on the track. Part B: B. Inelastic Collision : Glider I. m1 = 183.25 g Glider II. m2 = 283.05 g Trial 1 2 3 4 V1i (m/s) 0.268 m/s 0.253 m/s 0.261 m/s 0.271 m/s V(1 + 2) (m/s) 0.117 m/s 0.118 m/s 0.129 m/s 0.111 m/s V(1 + 2) from (equation 4) 0.105 m/s 0.0994 m/s 0.103 m/s 0.106 m/s Comments your results? Our results are precise. The average experimental velocity is 0.11875 m/s, while the average theoretical velocity is 0.10335 m/s. This means that we had an error of 15 %, which is not too much, making our results sufficiently accurate. The reason for this error, is because there is still friction on the track, the track is not 100% balanced and because the gliders themselves could have created some error in the readings, by being slightly deformed. Date: 15/11/2006 Physics 210L Exp. 1. Surface Tension & Viscosity A- Surface Tension measurements Radius of Aluminum ring R = 5.82 cm; Thickness of Ring = 1.0 mm. Weight of Ring W = 0.049 N i- Water Water temperature = 22.5 oC Trial number F (N) Surface tension force (N)  (N/m) Trial 1 0.0630 0.0140 0.0192 Trial 2 0.0632 0.0142 0.0194 Trial 3 0.0634 0.0144 0.0197 Trial 4 0.0633 0.0143 0.0196 Trial 5 0.0634 0.0144 0.0197 From this table, calculate the average value of water and its rms error water = 0.0195  0.0001 N/m ii -Cold Water Water temperature = 6.1 oC Trial number F (N) Surface tension force (N)  (N/m) Trial 1 0.0635 0.0145 0.0198 Trial 2 0.0631 0.0141 0.0193 Trial 3 0.0630 0.0140 0.0192 Trial 4 0.0633 0.0143 0.0196 Trial 5 0.0641 0.0151 0.0207 From this table, calculate the average value of water . water = 0.0197  0.0003 N/m iii- Soapy Water Trial number F (N) Surface tension force (N)  (N/m) Grade: Trial 1 0.0610 0.0120 0.0164 Trial 2 0.0611 0.0121 0.0166 Trial 3 0.0610 0.0120 0.0164 Trial 4 0.0605 0.0115 0.0157 Trial 5 0.0608 0.0118 0.0161 From this table, calculate the average value of  soapy water. soapy water = 0.0162  0.0002 N/m QUESTIONS: 1) According to your experimental data, which has the largest value of , water or soapy water? Is this to be expected? THE LARGEST SURFACE TENSION IS FOR THE WATER, AT ROOM TEMPERATURE. YES THIS IS TO BE EXPECTED. THIS IS SO SINCE SOAP IS SUPPOSED TO MAKE THE SURFACE TENSION LESS. THIS IS EXACTLY THE PRINCIPLE THAT DETERGENT USES TO CLEAN CLOTHES IN WASH MACHINES. IT DECREASES THE SURFACE TENSION SO THAT WATER PENETRATES THE CLOTHES EASIER, TO CLEAN THEM. 2) What is the effect of temperature on your measurements of surface tension? Comment. BASED ON OUR RESULTS, THE SURFACE TENSION INCREASES AS THE TEMPERATURE INCREASES. THIS IS SO SINCE AS THE TEMPERATURE DECREASES THE THERE IS LESS ENERGY IN THE SYSTEM. WHEN THE TEMPERATURE INCREASES HIGHER ENERGY ALLOWS MOLECULES TO OVERCOME THE FORCES BETWEEN ONE ANOTHER, MAKING THE LIQUID LESS VISCOUS. HOWEVER IF THE TEMPERATURE DECREASES THEN THERE IS LESS ENERGY AND THE MOLECULES ARE MORE ATTRACTED TO EACH OTHER. THERE WILL BE HIGHER FORCES OF ATTRACTION BETWEEN EACH OTHER. B- Viscosity Length and radius of capillary tube: l = 14.45 cm and R = 0.08 cm. h (cm) hf (cm) h(cm) Volume (cm3) Time (s) Temp. (oC) Q(cm3/s) (kg/ m.s) 8.18 7.97 8.08 12.0 9.9 22.7 1.21 0.00203 7.97 7.74 7.86 18.0 11.98 22.9 1.50 0.00162 8.28 8.19 8.24 8.00 4.38 22.9 1.83 0.00135 8.28 8.15 8.22 10.0 5.66 22.7 1.77 0.00140 8.28 8.16 8.22 9.00 5.45 22.7 1.65 0.00150 8.27 8.01 8.14 20.5 12.4 22.7 1.65 0.00150 8.21 7.89 8.05 24.0 14.91 22.7 1.61 0.00153 7.89 7.53 7.71 28.0 19.66 22.8 1.42 0.00170 From your results, calculate the average coefficient of viscosity and its root-mean- square error. How does your value compare with the accepted value of the viscosity of water? Calculate the relative error on your measured value.  = 0.00158  0.00008 0.00158 PA*S = 1.58 CP (CENTI-POISE) THE VISCOSITY OF WATER AT 200C IS 1.002 CP. THEREFORE OUR VALUE IS VERY CLOSE TO THE LITERATURE VALUE OF THE VISCOSITY OF WATER. OUR ERROR THEREFORE IS THE ABSOLUTE VALUE OF THE THEORETICAL VALUE MINUS THE EXPERIMENTAL VALUE. THEN DIVIDE THAT BY THE THEORETICAL VALUE AND MULTIPLY BY 100. SO THE ERROR = 57.7% THIS IS AN ACCEPTABLE VALUE. PHYSICS 210L Thermal Expansion of Solids Date: 29/11/2006 1.DATA: (a) Copper: L0 (mm) Rrm (k Trm (C) Rhot (k Thot (C) T (C) L (mm) 700 99.8 25 6.9 93 68 0.925 What are the values of: T = ± 1 C L = ± 0.005 mm L0 = ± 0.5 mm Estimate the value of  (for copper) and its rms error. Show your calculation of error propagation. Comment Copper = (19.4  0.3)*10-6 /C Calculate the percentage error of your measurement compared to the accepted value copper = 17.610-6 /C. WE HAVE AN ERROR OF 10.2 % THIS IS FOUND BY USING THE FORMULA OF PERCENTAGE ERROR. (b) Steel: L0 (mm) Rrm (k Trm (C) Rhot (k Thot (C) T (C) L (mm) 700 108.1 23 8 88.5 65.5 0.67 Estimate the value of  (for Steel) and its rms error. Show your calculation of error propagation. Comment Steel = (14.6  0.2)*10-6 /C Calculate the percentage error of your measurement compared to the accepted value steel = 12.410-6 /C. 1 Grade: IN THIS EXPERIMENT WE HAVE A PERCENTAGE ERROR OF 17.7%. ONCE AGAIN WE USED THE FORMULA FOR PERCENT ERROR TO FIND THE VALUE OF THE ERROR. (c) Aluminum: L0 (mm) Rrm (k Trm (C) Rhot (k Thot (C) T (C) L (mm) 700 97.3 25.5 8.5 88.5 63 1.21 Estimate the value of  (Aluminum) and its rms error. Comment Aluminum = (2.74  0.05)*10-5 /C Comment on the precision and accuracy of your measurements and speculate on the possible sources of error in your experiment. ALL OUR VALUES ARE ACCURATE. HOWEVER, WE ARE NOT SURE ABOUT THE LAST MEASUREMENT SINCE WE CAN NOT COMPARE IT WITH ANYTHING. BUT, SINCE THE OTHER ONES WERE VERY ACCURATE, AND SINCE WE FOLLOWED THE SAME STEPS IN THIS EXPERIMENT, ALL OUR VALUES MUST BE ACCURATE AND PRECISE. WE HAVE HIGH PRECISION BECAUSE THE ERRORS ON THE VALUES ARE LOW. OF COURSE OUR SOURCES OF ERROR ARE OBVIOUS. WE MAY HAVE ASSUMED THE WRONG TEMPERATURE WHILE LOOKING FOR IT ON THE SCALE OF COMPARISON. WE DIDN'T CALCULATE THE EXACT MAXIMUM OF THE L. THERE MAY HAVE BEEN ERRORS IN THE EQUIPMENT THAT WE USED. From your result, calculate the coefficients of volume expansion for copper, Steel and aluminum? β=3 βCopper = (5.82 0.3)*10-5 /C βSteel = (4.38 0.06)*10-5 /C βAluminum = (8.2 0.2)*10-5 /C 2 Date: 13/12/2006 Physics 210L Mechanical Equivalent of Heat Experimental procedure: Copper (g) Aluminum (g) Iron (g) Weight of the calorimeter and uncertainty on measurement 645.7 212.2 580.5 Measure the diameter (d) of each calorimeter 4 times. Calculate the average value and the root mean square error. Copper ( cm ) Aluminum ( cm ) Iron ( cm ) 4.64 4.65 4.66 4.63 4.66 4.65 4.65 4.64 4.66 4.64 4.65 4.66 Average: 4.640 4.650 4.658 Root mean square: 0.004 0.004 0.003 A- Copper Record the initial temperature Ti final temperature Tf and number of turns n. Record the weight attached: 5.00 Kg Ti Tf T n first reading 25.3 oC 33.3 oC 8.0 oC 300 oC second reading 33.3 oC 39.2 oC 5.9 oC 300 oC third reading 39.2 oC 47.1 oC 7.9 oC 300 oC Calculate the mechanical equivalent of heat in each case (show calculations) Find its average value and its root mean square error. J = 4.91 ± 0.5 J/kcal Compare the result to the literature value and comment. B- Aluminum Record the weight attached 5 kg and the number of turn n = 300 measure Ti = 24.2 oC & Tf = 35.9 oC Grade: 1 Find the specific heat of Aluminum and the percentage error. Calculate the molar heat capacity, (show calculation) C- Iron Record the weight attached 5 kg and the number of turns n = 300 Measure Ti = 25.2 oC & Tf = 33.6 oC Find the specific heat of iron, its rms error as calculated from the error propagation theory (show calculation) and its rms error. Calculate the molar heat capacity of Iron. Are your results for Aluminum and Iron in agreement with the law of Dulong and Petit? Comment. THE LAW OF DULONG AND PETIT STATES THAT THE MOLAR HEAT CAPACITIES OF SOLIDS, WITH A FEW EXCEPTIONS, HAVE VALUES CLOSE TO 6 CAL/(MOL. OC). THE RESULTS THAT WE GOT WERE VERY CLOSE TO THIS VALUE. THE MOLAR HEAT CAPACITY OF ALUMINUM WAS 5.319 CAL/(MOL. OC) AND THE MOLAR HEAT CAPACITY OF IRON WAS 5.72 CAL/(MOL. OC). THE PERCENT ERROR OF EACH ONE ARE AS FOLLOWS: Aluminum Percent Error: Iron Percent Error: THESE LOW ERRORS MEAN THAT OUR RESULTS WERE VERY ACCURATE. NOW WE CAN'T REALLY TALK ABOUT PRECISION SINCE PRECISION IS DEFINED OVER A SET OF READINGS, HERE WE ONLY HAD ONE READING. THERE ARE MANY SOURCES FOR THIS ERROR: EQUIPMENT PROBLEMS, US, NOT ENOUGH CONTACT BETWEEN THE ROPE AND THE CYLINDRICAL PIECE OF METAL, ETC… 2 Physics 210L Standing Sound Waves in Air Columns Grade: Tube length and resonant modes: i. Determination of piston positions that produce standing waves: Freq.= 1.0 kHz Freq.= 1.5 kHz Freq.= 2.0 kHz Freq.= 2.5 kHz Piston Positions (cm) Piston Positions (cm) Piston Positions (cm) Piston Positions (cm) 72.5 71.5 73.0 75.5 55.7 60.0 63.8 68.5 39.0 48.5 54.8 61.6 21.6 36.6 45.9 49.8 4.00 25.0 36.0 48.0 13.5 28.0 41.1 3.00 19.0 34.5 9.90 27.6 0.600 20.7 13.8 7.05 ii. Determination of the speed of sound in air: Frequency (kHz) 0.5 1.0 1.5 2.0 d (cm) 33.5 17.125 11.42 8.95 v (m/s) 335 342.5 342.5 358 The average measured value of the speed of sound in air, vrms, is: (345  5) m/s Compare the measured value of the speed in air to the theoretically accepted value. Show your calculations and estimate the relative error. 1 a- Standing wave pattern in an air column : i. Waveform in the tube: Frequency = 1.5 kHz Microphone Positions Maxima Minima 0.5 cm 5 cm 12 cm 16.2 cm 24.8 cm 28 cm 34 cm 39.5 cm 45 cm 51.1 cm 56 cm 62.5 cm 68 56 cm Sketch the variations of the displacement and the pressure amplitude along the tube length. From your sketch, can you tell whether the piston and the speaker are considered as displacement nodes or anti-nodes? Explain. The Varying of The Displacement Amplitude 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 Displacement (cm) Amplitude The Varying of the Pressure Amplitude 0 10 20 30 40 50 60 70 80 Displacement (cm) Amplitude 2 FROM OUR SKETCH WE CAN SEE THAT THE SPEAKER IS A DISPLACEMENT ANTI-NODE SINCE THERE IS A MAXIMUM AMPLITUDE AT THE BEGINNING OF OUR SKETCH. HOWEVER, SINCE THE TUBE HAS A FIXED END, THIS FIXED END DOES NOT ALLOW DISPLACEMENT MEANING THAT THE PISTON IS A DISPLACEMENT NODE. ii. Determination of the speed of sound in air: Record the number of pressure nodes, N, for each value of the tube length. Frequency = 1.5 kHz Length (cm) N 71.5 6 60 5 48.5 4 36.6 3 25 2 13.5 1 3 0 Plot the length of the tube as a function of the number of pressure nodes in the standing wave pattern. What is the slope of the straight line that you have obtained? Calculate from this slope the velocity of sound in air. xi yi y ei ei 2 xi2 0 0.03 0 -0.03 0.00090 0 1 0.135 0.1205 -0.0145 0.000210 1 2 0.25 0.241 -0.009 8.1*10-5 4 3 0.366 0.3615 -0.0045 2.03*10-5 9 4 0.485 0.482 -0.003 9.00*10-6 16 5 0.600 0.6025 0.00250 6.25*10-6 25 6 0.715 0.844 0.129 0.0166 36 ∑(ei)2 = 0.0178 (∑xi)2 = 441 ∑(xi)2 = 91 ∆ = 196 ∑xi = 21 THE SLOPE ERROR IS  0.0113 THE SLOPE FROM THE GRAPH IS  SLOPE = (0.1205  0.01) THE SLOPE IS EQUAL TO HALF THE WAVE LENGTH. THEREFORE THE VELOCITY IS EQUAL TO THE PRODUCT OF (2)(SLOPE)(FREQUENCY)= (2)(0.1205)(1500)  V=361.5 M/S 3 The Number of Nodes With Displacement y = 0.1205x R2 = 0.9965 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0 1 2 3 4 5 6 7 Number of Nodes Length of Tube (m) How does this value compare to the one determined in part (a) of the experiment and to the accepted theoretical value? Comment. THE SPEED OF SOUND FROM PART (A) AND THEORETICALLY ARE APPROXIMATELY EQUAL, THEREFORE THE VALUE OF THE SPEED OF SOUND IN THIS PART WILL HAVE A SIMILAR ERROR WITH THE VALUE OF PART (A). THE FIRST PERCENTAGE ERROR IS THE ERROR WITH THE THEORETICAL VALUE OF THE SPEED OF SOUND, WHILE THE SECOND PERCENTAGE ERROR IS THE ERROR WITH THE VALUE FROM PART (A). 4 Date: 20/12/2006 Physics 210L Michelson Interferometer 1. Data: Determination of the wavelength of the LASER: dm (m) N  (m) Eq. (1) 33.1 100 0.662 29.2 100 0.584 30.0 100 0.600 32.3 100 0.646 35.0 100 0.700 Calculate the average value of l and its rms error. Compare the expected value and comment. m Mean value) ±m b. Determination of the index of refraction of air: (l=3.00 cm and  = 632.8 nm) Trials N nair (Eq.5) 1 11 1.00012 2 12 1.00013 3 12 1.00013 4 12 1.00013 5 11 1.00012 Calculate the average of nair and its rms error. Compare to the expected value and comment. nair = 1.000126 ± 0.000002 c. Determination of the index of refraction of glass: Thickness of plate: 0.54 cm. 1 Grade: N  Nglass (Eq.6) 110 11o 1.53 160 12o 1.73 150 11.7o 1.72 147 11.3o 1.78 170 12.05o 1.81 Calculate the average of nglass and its rms error. Compare to the expected value and comment. Nglass = 1.714 I used this one glass, borosilicate (pyrex) 1.474 Nglass =1.71 ± 0.05 glass, crown (soda-lime) 1.512 glass, flint (29% lead) 1.569 glass, flint (55% lead) 1.669 glass, flint (71% lead) 1.805 glass, fused silica 1.459 1. In the calculation to determine the value of λ based on the micrometer movement, why was dm multiplied by two? THIS IS BECAUSE A WAVE HAS TWO ABSOLUTE MAXIMUMS WHERE ONE IS A MINIMUM AND THE OTHER IS A MAXIMUM. HOWEVER, THERE IS CONSTRUCTIVE INTERFERENCE AT BOTH OF THESE EXTREMAS. SO THE BRIGHT LIGHTS THAT WE SAW WERE FROM THE MAXIMUMS AND THE MINIMUMS OF EACH WAVELENGTH, MEANING THAT EVERY TWO BRIGHT LIGHTS CORRESPONDED TO ONE WAVELENGTH. 2. To what do you attribute this difference between the measured value of nglass & accurately known value of nglass? THERE ARE MANY REASONS FOR ERROR BUT OUR RESULTS SHOW THAT WHAT WE DID WAS FAIRLY ACCURATE. OUR READINGS WERE VERY PRECISE ALL THROUGH. NOW THE REASON FOR ERROR IN THE INDEX OF REFRACTION IN GLASS COULD BE TO THE FACT THAT THE GLASS ISN'T THAT PURE. AS YOU CAN SEE FROM THE VARIOUS INDEXES OF REFRACTIONS OF GLASS, EACH TYPE OF GLASS HAS A DIFFERENT COMPOSITION. SO THERE MAY BE IMPROPER MEASUREMENTS FOR THE COMPOSITION OF GLASS. THE GLASS MAY NOT BE AS CLEAR AS IT SHOULD BE WHICH WOULD REPEL SOME OF THE LASER BEAM'S RAYS. OF COURSE THE EQUIPMENT IN GENERAL MAY NOT BE ACCURATE. ALSO OUR READING OF THE MEASUREMENTS MAY NOT HAVE BEEN SO PRECISE BECAUSE EVERYTHING WAS DELICATE. SO THERE ARE MANY SOURCES OF ERROR FOR FINDING THE INDEX OF REFRACTION OF GLASS. 2