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# EXPERIMENTAL PHYSICS 1 PHYS 1140

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This 13 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 1140 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/232108/phys-1140-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.

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Date Created: 10/30/15

M5l Lab M5 Hooke s Law and the Simple Harmonic Oscillator Most springs obey Hooke s Law which states that the force exerted by the spring is proportional to the extension or compression of the spring from its equilibrium length a F kx displacement X k is called the spring constant and is a measure of the H stiffness of the spring The minus sign indicates that the direction of the force is opposite to the direction of mass m the displacement In the SI system the units of spring M constant k are Nm The diagram shows a mass to the WWW right of its equilibrium position X0 The displacement of the mass is and the spring force is X0 to the left If the spring is compressed then X is and the spring force is W Simple harmonic motion occurs whenever there is a restoring force which is proportional to the displacement from equilibrium as is the case here Assuming no frictional forces and assuming that the spring is massless the equation of motion ma F ofa mass on a spring is dzx dzx k m kX or X 2 dt2 dt2 m The solution of this secondorder differential equation is k 3 Xt A s1n03tltlgt 03 m where A and I are constants which depend on the initial conditions the initial position and initial velocity of the mass The period T the frequency f and the constant 03 are related by 211 4 2 f n 11 T Thus the period T is given by m 5 T21T k A very important property of simple harmonic motion is that the period T does not depend on the amplitude of the motion A Fall 2004 M52 If the mass is hung from a vertical spring it will still execute simple harmonic motion with the same period 5 as we will now show When the mass is hung from the spring the spring is stretched from its equilibrium length by the weight mg of the mass The equilibrium displacement of the mass under the in uence of the force of gravity down and the y0 force from the spring up is Xequ Now that there are two forces acting on the y X mass the equation of motion becomes 2 d 6 m ZX mg k x dt We now perform a change of variable by introducing a new coordinate y x 2 2 d d Since 1s a constant differentiating y tw1ce produces 3 Also k dt dt mg kx ky therefore eq n 6 becomes 2 7 min 2y ky which is exactly the same as eq n 2 except we have changed the name of the position coordinate from x to y Since the equation of motion is the same the solution is the same 3 with the same period 5 A real spring has mass a fact which we have ignored so far A mass m on a real spring with mass mspring oscillates more slowly than predicted by 5 since the spring has to push its own mass about as well as the mass m However the theoretical expression 5 can be corrected by replacing the mass m with an effective mass consisting of m plus some fraction of mspring 8 Ineff m f 39 Inspring where f is some fraction f lt l which depends on the exact shape of the spring For the spring used in this lab the fraction f has been determined experimentally to be 9 f 0281i0002 Fall 2004 M53 Although f can be computed the computation is rather complicated and depends on the precise shape of the spring In this experiment you will first determine the spring constant k of a spring by hanging various weights from the spring and measuring the extension Then you will measure the period of oscillation of a mass m hanging from the spring and will compare this measured period with period predicted by eq n 5 with meff used instead of m Finally you will repeat your measurement of the period with two other masses and check that the period is proportional to 1 meff Procedure Part 1 Measurement of the spring constant k Begin by weighing the spring you will need the weight in part 2 You might want to check the reliability of the digital balances by weighing the spring on two different balances Use the digital balances with 0 lg resolution The spring used in this lab has a tapered coil which serves to reduce the interference of the windings with each other and make the spring behave more like a perfect Hooke s spring Always hang the spring with the larger windings downward With the empty mass tray hanging from the spring measure the position of an edge of the tray This will be the zero position which you will subtract from all subsequent positions There is a mirror by the meter stick scale so you can avoid parallax when you measure positions Now add the slotted weights to the tray one at a time and measure the positions of the same edge as each mass is added Add masses in 50 gram increments to atotal of 500 grams Am 50 g 100 g 150 g Use the balance to check that the masses are accurate Plot the weight added Am g vs Ax the position change from the zero position Use the known value of g 9796 msz This graph should be a straight line with slope k tapered spring mass tray To determine the slope compute k Am g for each data pomt and compute x the mean the standard deviation and the standard deviation of the mean of your several k values iki kavg 2 6 6 T mean W 1 N 1an 2ki 6 i l Side comment You might think to determine the best value of k by measuring the slope of Am g vs Ax using the slope function in Mathcad This is not quite right Fall 2004 M54 since what we want is the best fit line which goes through the origin while the slope function nds the slope of the best fit line which in general has a nonezero intercept Part 2 Measurement of the period Remove the weight tray from the spring and load the spring with a 100 gram mass with a hook Carefully set the mass oscillating with a small amplitude motion and use a stop watch to time the interval for several complete oscillations If the time for N complete T periods N m 50 or more is Tum then the period is Tmea1S ml mass with hook By measuring the time for many periods the uncertainty in T due to your reaction time is reduced by a factor of N 5T 5Ttota1 human reaction time u 01 sec 5T t Oml DO NOT measure one period 50 times measure the time for 50 periods once If you have time repeat this measurement at least once to check the reproducibility of this method Now compute the period using eq n 5 with meg in place ofm 10 Teak 21 1111 Also compute 5Tca1c the uncertainty in Teak see note on uncertainty calcs below Compare the calculated and measured periods Repeat this procedure with m 200g and m 500g but you need not repeat the calculation of 5Tca1c it s too timeconsuming Make a plot of Tmeals vs 211 Eff using your three data points On the same graph plot the line y X Finally compute T the quantity R Tmeas for each of your three data p01nts Do theory and experiment calc agree 1391 In Review For a function like f X yz X involving multiplication division and 5f 5x 2 5y 2 52 2 powers the fractional uncertainty is given by n m p X y 2 Fall 2004 M55 PreLab Questions 1 What condition or conditions are necessary for simple harmonic motion to occur 2 What are the units of the spring constant k in SI units which are also called MKS units MKS meterkilogramseconds What are the units in the cgs system centimetergramsecond 3 A given spring stretches 15 cm when it is loaded with a 250 gram weight What is the spring constant of the spring 4 Sketch the graph vs AX for a mass on a spring Indicate the slope and intercept 5 What happens to the period of a massonaspring simple harmonic oscillator if the mass is doubled What happens to the period if the spring constant is reduced by a factor of 3 How does the period depend on the amplitude of the oscillation 6 True or False for a simple harmonic oscillator consisting of a mass on a spring the period measured when the mass is hanging vertically gure on p2 is different than the period measured when the spring and mass are supported horizontally figure on p l 7 Verify that eq n 3 is a solution of the differential equation 2 Hint There was a very similar problem in the Simple Pendulum Lab Ml 8 Show with a sketch what a graph of T2 vs m ff looks like What are the slope and intercept 9 Assuming that k meff 5k and 5 meff are known show how to compute 5Tca1c where Teak is given in eq39n10 10 A mass m hanging from a spring with spring constant k is taken to the Mars where g is 13 of its value on Earth and the period of the oscillation is measured Is the measured period on the Mars the same as the period measured on the Ealth Explain your answer Fall 2004 E3l Lab E3 The Wheatstone Bridge Introduction The Wheatstone bridge is a circuit used to compare an unknown resistance with a known resistance The bridge is commonly used in control circuits For instance a temperature sensor in an oven often consists of a resistor with a resistance that increases with temperature This temperaturedependent resistor is compared with a control resistor outside the oven to control a heater and maintain a set temperature A schematic of a Wheatstone bridge is shown below Fig1 Schematic of a Wheatstone Bridge The unknown resistor is Rx the resistor Rk is known and the two resistors R1 and R2 have a known ratio R2 R1 although their individual values may not be known A galvanometer G measures the voltage difference VAB between points A and B Either the known resistor Rk or the ratio R2 R1 is adjusted until the voltage difference VAB is zero and no current ows through G When VAB 0 the bridge is said to be balanced Since VAB 0 the voltage drop from C to A must equal the voltage drop from C to B VCA VCB Likewise we must have VAD VBD So we can write Ia R1 1b Rk 2 Ia R2 1b Rx Dividing 2 by l we have R R R 3 2 X Rx Rk 2 R1 Rk R1 Fall 2004 E32 Thus the unknown resistance Rx can be computed from the known resistance Rk and the known ratio R2 R1 Notice that the computed Rx does not depend on the voltage V0 hence V0 does not have to be very stable or wellknown Another advantage of the Wheatstone bridge is that because it uses a null measurement VAB 0 the galvanometer does not have to be calibrated In practice the Wheatstone bridge is seldom used merely to determine the value of a resistor in the manner just described Instead it is usually used to measure small changes in Rx due for instance to temperature changes or the motion of microscopic defects in the resistor As an example suppose Rx 106 Q and we wanted to measured a change in Rx of 19 resulting from a small temperature change There is no ohmmeter which can reliably measure a change in resistance of 1 part in a million However the bridge can be set up so that VAB 0 when Rx is exactly 106 Q Then any change in Rx ARX would result in a nonzero VAB which as we show below is proportional to ARX You would not weigh a cat by weighing a boat with and without a cat on board Likewise you would not want to measure very small changes in Rx by measuring Rx with and without the change Instead you want to arrange things so that the change in Rx ARX is the entire signal The bridge serves to balance out the signal due Rx leaving only the signal due to ARX To show that VAB cx ARXwe consider Fig1 and note that VCD V0 We assume that the galvanometer is a perfect voltmeter so that no current ows through it even when the bridge is not balanced We also assume that the bridge has been balanced with the sample resistance at an initial value of Rxo so that RXO RkR2R1 Then we consider what happens to VAB when the sample resistor is changed by a small amount to anew value RX RXO ARK Applying Kirchhoff s Voltage Law and Ohm s Law to the upper and lower arms of the bridge we have 4 Vo1aRlR2IbRk Rx We are trying to nd VAB which we can relate to Ia and lb 5 VABVA VBVC VB VC VA1bRk 1aR1 We can use 4 to solve for Ia and lb and then substitute for Ia and lb in 5 R R 6 VAB Vo kVo 1 Rk RX R1 R2 This equation shows how VAB depends on Rx Notice that eqn6 yields VAB 0 when RX RkR2R1 To see how much VAB changes when Rx changes from Rx0 to RXO ARK we write Fall 2004 E33 dV R AB ARX V0 dRX Rx Rx0 Rk Rxo RX 7 AVAB E We wrote eqn7 by regarding VAB as a function of Rx and remembering from Calculus that if f fX then of 3 f395X X Substituting RXO RkR2R1 into 7 yields 8 2 V R V V R Awm LJLARX L mx 27 R z R2 RkRk R1 RklR1 Finally we remember that VAByinitial 0 so the change in VAB is the same as VAB and we have 2 R AR 9 VAB X 1R2 Rk We are done We have shown that when the bridge is balanced any small change in Rx will produce a VAB proportional to that change Experiment We will use a slidewire Wheatstone bridge in which the two resistors R1 and R2 are two portions of a single uniform NiCr wire Electrical contact is made at some point along the wire by a sliding contact this contact corresponds to point A The two portions of the wire on either side of the contact have resistances R1 and R2 and the ratio R2 R1 is the same as the ratio of the lengths of the two portions of wire L2 L1 The lengths are readily measured with a meter stick which the wire rests upon A digital multimeter DMM in voltage mode will serve as the galvanometer Fall 2004 E34 10 Q resistor 6 V power supply Rx unknown Rk decade box B 39 NiCr wire on meter stick A sliding contact L1 L2 DMM Fig 2 Physical Layout of the Wheatstone Bridge The 109 resistor in series with an adjustable power supply serves to limit the current through the bridge to less than lA Higher currents might overheat components of the bridge Set the voltage from the power supply to approximately 6 volts with the Coarse adjust voltage knob The known resistor Rk is adjustable and can be set to any value from IQ to 9999 in IQ steps with a decade resistance box which is accurate to 002 The NiCr wire has a total resistance of about 29 The sliding contact is spring loaded you have to push it down to make contact to the wire at point A There are two buttons on the sliding contact push one or the other but not both to contact the wire Wire to sliding contact sliding contact LL l39I GI nichrome wire Adjust the position of the sliding contact to balance the bridge zero reading on the DMM For Part 1 of the lab the unknown resistor Rx is one of 5 coils of wire numbered 1 through 5 mounted on a board The lengths of the wires their composition and their Fall 2004 E35 gauge number are printed on the board The gauge number is a measure of the thickness of the wire 22 gauge wire has a diameter of 0644 mm 28 gauge wire has a diameter of 0321 mm For part 2 ofthe lab Rx is a standalone coil of copper The resistance R of a wire is related to its length L its crosssectional area A and its resistivity p by the formula L 10 R p A The resistivity p of a material depends on composition on defect structure and on the temperature For metals p is approximately constant at very low temperatures T lt 100K and increases approximately linearly with temperature measured in K at high temperatures At T 20C the resistivity of pure defectfree copper is pRT 1678 gtlt1078 Q m RT stands for room temperature RT293K T 0K Procedure Part 1 Resistivity of Copper In this section you will use the bridge to make precise measurements of the resistance of each of the 5 coils of wire on the coil board First however use the digital multimeter DMM to make an approximate measurement of the resistance of each coil Also use the DMM to see how the resistance ofthe decade box changes when you turn the knobs You must temporarily disconnect the coil board and the decade box from the bridge when testing them with the DMM With the DMM measure the resistance of each of the ve coils to the nearest 01 Q The resistance ofthe wire leads used to connect the DMM to the coils is not negligible so first use the DMM to measure the resistance of the two wire leads in series and then subtract this lead resistance from your measurements Check yourself the smallest coil resistance is below 1 ohm Now familiarize yourself with the decade resistance box Adjust the knobs for 15 ohms and verify with the DMM that the value is indeed 15 ohms plus the resistance of Fall 2004 E36 the wire leads For this measurement disconnect the decade box from the bridge circuit and connect the DMM directly to the decade box Now use the bridge to measure the resistance of each of the 5 coils Before connecting the power supply to the bridge carefully check that all the connections are correct Select a coil attach it to the bridge and set the decade box resistance Rk to be as near to Rx as possible you know Rx roughly from your DMM measurements Balance the bridge by moving the sliding contact along the wire while watching the DMM Remember that when using the galvanometer you should always begin with button 1 then button 2 and finally button 3 With the bridge balanced measure L1 and L2 and compute 11 R R RZR L2 x le kLl Repeat this procedure for the other 4 coils For each of these coils compute the resistivity using your measured resistances the data printed on the coil board and eq n 10 Make a table with the headings coil R from DMM R from bridge and computed resistivity Four of the 5 sample coils are made of copper Cu For these four coils compute the average resistivity and the uncertainty of the average smean Compare your average value with the known value Coil 5 is made of a coppernickel alloy What is the ratio pcuNi pcu Part 2 Temperature dependence of resistivity For this part the unknown Rx is a copper wire coil the one which is not attached to the coil board We will use eq n9 to measure the change in resistance of the coil when its temperature is changed by plunging it into an ice bath Note that eq n9 can be written as 2 L AR 9 VAB V0 L 1 X 1L2 Rk Prepare an ice bath by filling a beaker with ice from the freezer ask your TA where and adding some water Measure room temperature and the ice bath temperature with the digital thermometer Measure the coil resistance Rx with the DMM and set Rk equal to Rx as nearly as possible With RK and Rk connected in the bridge measure VO VCD with the DMM See Fig 2 to locate VCD Note that VCD is not the power supply voltage For this bridge measurement we must replace the galvanometer in the bridge circuit with the DMM set to DC volts with the most sensitive range400mV We need to do this because the galvanometer does not behave as an ideal voltmeter as assumed in the derivation of eq n 9 Fall 2004 E37 With the sample coil at room temperature balance the bridge and compute the coil s resistance at room temperature Now without changing the position of the slide wire contact place the coil into the ice bath and watch as the voltage VAB on the DMM changes Record VAB after it reaches equilibrium Use eq n9 to compute ARX the change in resistance ofthe coil As a measure of how sensitive the resistivity is to temperature changes we can compute the fractional change in the resistivity divided by the change in temperature A 12 Ml iamp AT pAT R AT This quantity multiplied by 100 is the change per degree From your measurements compute the change in resistivity per degree for copper and compare this value with the value you expect if p cx T K Hint If p cx T K then p CT where C is some constant and Ap C AT Fall 2004 E38 Prelab Questions 1 Two copper wires labeled A and B are at the same temperature but the temperature is unknown Wire B is three times as long and has 13 the diameter of wire A Compute the ratio RBRA 2 A 28 gauge copper wire is 10 meters long What is its resistance at room temperature 3 Name two advantages of a Wheatstone bridge over an ordinary ohmmeter 4 What is the formula for the total resistance of two resistors in parallel Consider two resistors R1 2 9 and R2 200 9 What is the total resistance of these two resistors in parallel Give your answer to the nearest ohm 5 Consider the circuit shown in Fig2 not Fig1 and assume that the galvanometer acts like an ideal voltmeter with in nite internal resistance Recall that the total resistance of the NiCr wire is 29 If Rk and RK are both very very large compared to 29 how much current ows through the NiCr wire 6 Suppose an unknown resistance Rx is measured with the bridge circuit shown in Fig 2 and the result is Rx 8659 The 6V power supply is then replaced with an 10V supply and RK is remeasured What is the new measured value of Rx 7 A sample of copper wire is 200m long and has a diameter of0 150 mm Its resistance is determined to be 8809 What is the resistivity of the copper in this wire 8 Counts as 3 questions A Wheatstone Bridge such as in Fig 1 has a V0 100V With Rk 5009 and the unknown resistor Rx at room temperature the bridge is balanced with L1 340cm and L2 660 cm What is the value of Rx While still attached to the bridge the unknown Rx is placed in an oven which raises its temperature by 3300 C The value of VAB is then found to be 0127V What is the resistance change of the unknown ARK What is the change per degree of the resistivity of this sample Fall 2004

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