《基础物理实验 Mental Physics Laboratory》课程教学资源：实验指南 Lab Guides 2018

Lab guides 2018 Homework Lab I Target hitting through collision Lab 2 Torsional pendulum Lab 3 The latent heat of vaporization of liquid nitrogen........ 13 Lab 4 Digital Storage Oscilloscopes 18 ab 5 Wheatstone bridge Lab 6 Magnetic field variation along the axis of a circular coil and a Helmholtz coil Lab7 Converging Lens…… 32 Lab 8 Frank-Hertz experiment with Neon 36 Lab 9 X-ray experiment Fundamental Physics Laboratory, Fudan University http://phylab.fudan.edu.cn

Lab Guides 2018 Homework 1 Lab 1 Target hitting through collision 2 Lab 2 Torsional pendulum 7 Lab 3 The latent heat of vaporization of liquid nitrogen 13 Lab 4 Digital Storage Oscilloscopes 18 Lab 5 Wheatstone bridge 23 Lab 6 Magnetic field variation along the axis of a circular coil and a Helmholtz coil 27 Lab 7 Converging Lens 32 Lab 8 Frank-Hertz experiment with Neon 36 Lab 9 X-ray experiment 40 Fundamental Physics Laboratory, Fudan University http://phylab.fudan.edu.cn

Home work 1. Please rewrite the following expression correctly 1)38.17±0.05322)1.1762±0.013)85300±200 2. Find out the final results(Please give the detailed procedures) 1)3.00*2.20+450*1.00+20*0.1 0.10002(0.5000 (Where the"1"is accurate. 3)26.80*103-20 (2480-220)*5989 2.00 5)2.150*3000*(1+m)(Where the"I"is accurate but the"4" has one significant figure. 205 3. Measure the inner diameter of a cylinder with a vernier caliper for 10 times, which has its maximum uncertainty a 0.002cm. The data is listed in the table dcm23222.3302.3242.318232223262.318232423262326 Please calculate the average value of d and its uncertainty 4. We measure the length of an object with a ruler, which has the smallest scale Imm and its maximum uncertainty 0.10 mm. If the parallax is well eliminated The readout of the right end is 1F17 52cm and that of the left end is 12=5.00cm. Please find the length / and its uncertainty u(y Express the final results in the form of ltu( 5. If the H±u(H)=700±0.02,G±uG)=(3.50±0.01)*10 H a) Ifx=-, please find x tu(x) G H b) hyG F (where F=150, is accurate), please find y tu(y) 6. During the measurement of the gravity acceleration g through a pendulum, we got data as listed in the table below ength L/cm.4 61.5 81.0 Period T/2|2.1832468 3.262 3.618 3.861 Please plot the data in a graph paper, calculate the slope of the line obtained and the gravity acceleration

1 Home work 1. Please rewrite the following expression correctly. 1) 38.17  0.0532 2) 1.1762  0.01 3) 85300  200 2. Find out the final results (Please give the detailed procedures) 1) 3.00*2.20+45.0*1.00+20*0.1 2) 2 2 (0.5000) 1 (0.1000) 1  (Where the "1" is accurate.) 3) 26.80*103 – 20 4) 2.00 (2.480 2.20)*5.989 5) 2.150*3.000*（1+ 4 205 ）(Where the "1" is accurate, but the "4" has one significant figure.) 3. Measure the inner diameter of a cylinder with a vernier caliper for 10 times, which has its maximum uncertainty a 0.002cm. The data is listed in the table: No. 1 2 3 4 5 6 7 8 9 10 d/cm 2.322 2.330 2.324 2.318 2.322 2.326 2.318 2.324 2.326 2.326 Please calculate the average value of d and its uncertainty. 4. We measure the length of an object with a ruler, which has the smallest scale 1mm and its maximum uncertainty 0.10 mm. If the parallax is well eliminated. The readout of the right end is l1=17.52cm and that of the left end is l2=5.00cm. Please find the length l and its uncertainty u(l). Express the final results in the form of l  u(l)=___________. 5. If the H u H    ( ) 7.00 0.02, 2 G u(G) ( . . )* 3 50 0 01 10    , a) If x= G H , please find x u(x ) ; b) If y= G H – F, (where F=150, is accurate), please find y u(y ). 6. During the measurement of the gravity acceleration g through a pendulum, we got data as listed in the table below. Length L/cm 53.4 61.5 71.2 81.0 89.5 95.5 Period T 2 /s2 2.183 2.468 2.877 3.262 3.618 3.861 Please plot the data in a graph paper, calculate the slope of the line obtained and the gravity acceleration

Lab 1-Mechanics laboratory ---Target hitting through collision Goal This experiment is designed for the review of mechanics. In this experiment, learn to predict, nd ther Related topics Projectile motion, collision, conservation law Introduction motion after being collided by another steel ball of the same mass at the bottom of its Swing. In this experiment, you will be challenged to hit the target with a steel ball that is in projectile The followings are definitions of some related concepts Projectile motion is the motion of an object moving through the air, subject to the acceleration of gravity. The object is called a projectile, and its path is called its trajectory o A conservative force is one, like gravity, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken Conservation of mechanical energy: the total mechanical energy (i.e. sum of kinetic energy and potential energy) of a system that experiences only conservation forces is constant, can only changes forms among kinetic energy and various types of potential energy. Such a system is called a closed syster Conservation of momentum: the total momentum of any isolated system, with any numbers of bodies in it, is conserved An isolated system is defined to be one for which the net external forces is zero o An elastic collision is one that conserves internal kinetic energy. Internal kinetic energy is the Im of the kinetic energies of the bodies in the sys An inelastic collision is one in which internal kinetic energy changes. This means the forces between colliding objects may remove or add internal kinetic energy For more details, please refer to your text book "College Physics", especially $3.4, $6.4-6 and§7.3-6 Strike BalI Struck ball d justable holder there is a horizontal line in Adjustable Stage the middle of the transparent box, which ndicates the height of the center of the strike ball Fig. I Schematic diagram of the experiment device

2 Lab 1 —Mechanics laboratory ——Target hitting through collision Goal This experiment is designed for the review of mechanics. In this experiment, learn to predict, operate, observe, analysis, and then improve...... Related topics Projectile motion, collision, conservation law Introduction In this experiment, you will be challenged to hit the target with a steel ball that is in projectile motion after being collided by another steel ball of the same mass at the bottom of its swing. The followings are definitions of some related concepts:  Projectile motion is the motion of an object moving through the air, subject to the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.  A conservative force is one, like gravity, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken.  Conservation of mechanical energy: the total mechanical energy (i.e. sum of kinetic energy and potential energy) of a system that experiences only conservation forces is constant, can only changes forms among kinetic energy and various types of potential energy. Such a system is called a closed system.  Conservation of momentum: the total momentum of any isolated system, with any numbers of bodies in it, is conserved. An isolated system is defined to be one for which the net external forces is zero.  An elastic collision is one that conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the bodies in the system.  An inelastic collision is one in which internal kinetic energy changes. This means the forces between colliding objects may remove or add internal kinetic energy. For more details, please refer to your text book "College Physics", especially §3.4, §6.4-6, and §7.3-6. Fig. 1 Schematic diagram of the experiment device There is a horizontal line in the middle of the transparent box, which indicates the height of the center of the strike ball

Experiment device Figure I is the sketch of the experiment device. During the experiment, very careful adjustment is necessary and helpful. Meanwhile, one should observe and analyze the forthcoming phenomenon roundl Procedure 1. Place the target-sheet in an appropriate position, the centerline should coincide with the centerline of the bottom plate 2. Place the projectile ball in its position 3. Adjust the system so that after release, the ball will move in the vertical plane as decided by the projectile ball and along the center of the target-sheet 4. Measure the position of the center of the target-sheet. Calculate at which height the swing ball would be left free, so that the projectile ball will hit the center of the target-sheet 5. Try the experiment once to verify your calculation 6. If the center of the target-sheet is not hit, analyze the possible reason(s) and recalculate. Try again to verify the new calculation 7. Procedure 6 may be repeated for several times until your "score"is better than gun circle Log your measured data, calculation, operation, phenomena and result in detail. Discuss the ossible reasons which may result in energy lost in this experiment. you have time, try this procedure again with a projectile ball of different mass, size, ol uestions 1. What is the main cause for energy lost in In this experI 2. Write down the related equations you needed in this experiment. Specify the ranges of their application References 1. Paul Peter Urone, College Physics, Chinese Machine Press, Beijing 2002 2. Physics 8A--Fall 2000 Lab Manual, University of California at Berkley 3.沈元华,陆申龙基础物理实验( Fundamental Physics Laboratory)高等教育出版社北京 2004pp.29-32

3 Experiment device Figure 1 is the sketch of the experiment device. During the experiment, very careful adjustment is necessary and helpful. Meanwhile, one should observe and analyze the forthcoming phenomenon roundly. Procedure 1. Place the target-sheet in an appropriate position, the centerline should coincide with the centerline of the bottom plate. 2. Place the projectile ball in its position. 3. Adjust the system so that after release, the ball will move in the vertical plane as decided by the projectile ball and along the center of the target-sheet. 4. Measure the position of the center of the target-sheet. Calculate at which height the swing ball would be left free, so that the projectile ball will hit the center of the target-sheet. 5. Try the experiment once to verify your calculation. 6. If the center of the target-sheet is not hit, analyze the possible reason(s) and recalculate. Try again to verify the new calculation. 7. Procedure 6 may be repeated for several times until your “score” is better than 9 th circle (inclusive). Log your measured data, calculation, operation, phenomena and result in detail. Discuss the possible reasons which may result in energy lost in this experiment. Optional If you have time, try this procedure again with a projectile ball of different mass, size, or composition. Questions 1. What is the main cause for energy lost in this experiment? 2. Write down the related equations you needed in this experiment. Specify the ranges of their application. References 1. Paul Peter Urone, College Physics, Chinese Machine Press, Beijing 2002 2. Physics 8A -- Fall 2000 Lab Manual, University of California at Berkley 3. 沈元华，陆申龙 基础物理实验 (Fundamental Physics Laboratory) 高等教育出版社 北京 2004 pp. 29-32

Operation guide--Lab 1 Mechanics laboratory Formulas derivation swing collision Projectile Figl: Schematic diagram a)ho: Initial height of the strike-ball b) m: Mass of the strike ball c)m: Mass of the struck ball d) D: Diameter of the ball, which should be measured for 5 times. Please read out from the vernier calipers correctly. e)y: Height of the adjustable stage fx: Distance between the center of the target sheet and that of the adjustable stage Please answer the followin 1) The initial gravitational potential energy of strike ball: Eo 2) Velocity of strike ball at the bottom of its swing vI= 3)Elastic Collision Conservation of energy Velocity of struck ball after being hit: v2= If m=2, rewrite the velocity of struck ball: v2= 4)Horizontal projectile motion Expression of drop height y and the distance x(which are functions of velocity v2 and time 0 :y= 5) The values of x, y, mI, and m can be measured easily in the lab. Please write the a height of the strike ball by using the above quantities(x, y, ml, and m2) If m=m2, rewrite the initial height: ho= 6) Since total energy of the system(strike ball struck ball) will be lost in the

4 Operation guide——Lab 1 Mechanics laboratory 1. Formulas derivation a) h0: Initial height of the strike-ball. b) m1: Mass of the strike ball. c) m2: Mass of the struck ball. d) D: Diameter of the ball, which should be measured for 5 times. Please read out from the vernier calipers correctly. e) y: Height of the adjustable stage. f) x: Distance between the center of the target sheet and that of the adjustable stage. Please answer the following questions: 1) The initial gravitational potential energy of strike ball: E0= . 2) Velocity of strike ball at the bottom of its swing: v1= . 3) Elastic Collision: Conservation of momentum: . Conservation of energy: . Velocity of struck ball after being hit: v2= . If m1=m2, rewrite the velocity of struck ball: v2= . 4) Horizontal projectile motion: Expression of drop height y and the distance x (which are functions of velocity v2 and time t): y = , x = . 5) The values of x, y, m1, and m2 can be measured easily in the lab. Please write the initial height of the strike ball by using the above quantities (x, y, m1, and m2): h0= . If m1=m2, rewrite the initial height: h0= . 6) Since total energy of the system (strike ball & struck ball) will be lost in the Fig1: Schematic diagram

actual experiment, the actual displacement of struck ball x is al ways smaller tha the distance x. Similar with the theoretical expression of initial height ho, you can easily write down the theoretical expression of h' which corresponds to the The lost energy in this trial is: AE= 7) In order to compensate the energy loss in previous trial, we should increase the initial gravitational potential energy of strike ball, in other words, the start height of the strike ball should be raised a little: 4H= 8)Procedure 7)may be repeated for several times until your " score "is better than 9th circle(inclusive)

5 actual experiment, the actual displacement of struck ball x’ is always smaller than the distance x. Similar with the theoretical expression of initial height h0, you can easily write down the theoretical expression of h’ which corresponds to the displacement x’: h’= . The lost energy in this trial is: ∆E’= . 7) In order to compensate the energy loss in previous trial, we should increase the initial gravitational potential energy of strike ball, in other words, the start height of the strike ball should be raised a little: H= . 8) Procedure 7) may be repeated for several times until your “score” is better than 9 th circle (inclusive)

2. Data Record 1)Do some adjustments in order to make sure your strike ball and struck ball had a head-on collision 2) Try the experiment once to verify your calculation 3)If the center of the target-sheet is not hit, analyze the possible reason(s)and make a new calculation. Readjust the device if necessary, especially the start height of the strike ball. (Write down the details of your observation, alculation and adjustment. ) Try again to verify the new calculation Data table No. 1: (Pay attention to the significant figures of your values. D/cm m/g m/g y/cm x/cm ho( Calculated)/em Data table no. 2: Shot position Possible reasons for the AEJJ Ho/cm No. Score 4H,/em (x', z")/cm ' miss-hit (average) H1= Shot position Possible reasons for the AEJ N Score 4H2/cm Ho+AHI x,4)/cm smiss-hit' avera Shot position Possible reasons for the AEvJ No. Scor 4H3/cm H1+4H2 (x', z)/cm smiss-hit' (average) Total energy loss: Ae= zis the distance of the shot position to the middle line of the target paper. It should a smal value 6

6 2. Data Record 1) Do some adjustments in order to make sure your strike ball and struck ball had a head-on collision. 2) Try the experiment once to verify your calculation. 3) If the center of the target-sheet is not hit, analyze the possible reason(s) and make a new calculation. Readjust the device if necessary, especially the start height of the strike ball. (Write down the details of your observation, calculation and adjustment.) Try again to verify the new calculation. Data table No. 1: (Pay attention to the significant figures of your values.) D/cm m1/g m2/g y/cm x/cm h0(Calculated)/cm Data table No. 2： z' is the distance of the shot position to the middle line of the target paper. It should be a small value. H0/cm No. Score Shot position (x', z')/cm Possible reasons for the "miss-hit" E1/J (average) H1 /cm 1 2 3 H1 = H0+H1 No. Score Shot position (x', z')/cm Possible reasons for the "miss-hit" E2/J (average) H2 /cm 1 2 3 H2 = H1+H2 No. Score Shot position (x', z')/cm Possible reasons for the "miss-hit" E3/J (average) H3 /cm 1 2 3 Total energy loss: E=

Lab 2 Torsional pendulum Goal This experiment is designed for a review of the rotation of rigid body Related topics Rotational motion, Oscillatory motion, Elasticity Introduction A torsional pendulum, or torsional oscillator, consists usually of a disk-like mass suspended from a thin rod or wire(see Fig. 1). When the mass is twisted about the axis of the wire, the wire exerts a torque on the mass, tending to rotate it back to its original position. If twisted and released the mass will oscillate back and forth, executing simple harmonic motion. This is the angular version of the bouncing mass hanging from a spring. This gives us an idea of moment of inertia We will measure the moment of inertia of several different shaped objects. As comparison, these moment of inertia can also be calculated theoretically. We can also verify the parallel axis theorem Given that the moment of inertia of one object is known, we can determine the torsional constant Fig. I Schematic diagram of a torsional pendulum This experiment is based on the torsional simple harmonic oscillation with the analogue of displacement replaced by angular displacement e, force by torque M, and the spring constant by torsional constant K. For a given small twist 6(sufficiently small), the experienced reaction is given by M=-K8 (1) This is just like the Hooke's law for the springs. If a mass with moment of inertia / is attached to the rod, the torque will give the mass an angular acceleration a according to m=a=/ get the following d-e K 6 Hence on solving this second order differential equation we get 8=Acos(t+o)

7 Lab 2 Torsional pendulum Goal This experiment is designed for a review of the rotation of rigid body Related topics Rotational motion, Oscillatory motion, Elasticity Introduction A torsional pendulum, or torsional oscillator, consists usually of a disk-like mass suspended from a thin rod or wire (see Fig. 1). When the mass is twisted about the axis of the wire, the wire exerts a torque on the mass, tending to rotate it back to its original position. If twisted and released, the mass will oscillate back and forth, executing simple harmonic motion. This is the angular version of the bouncing mass hanging from a spring. This gives us an idea of moment of inertia. We will measure the moment of inertia of several different shaped objects. As comparison, these moment of inertia can also be calculated theoretically. We can also verify the parallel axis theorem. Given that the moment of inertia of one object is known, we can determine the torsional constant K. Fig. 1 Schematic diagram of a torsional pendulum This experiment is based on the torsional simple harmonic oscillation with the analogue of displacement replaced by angular displacement , force by torque M, and the spring constant by torsional constant K. For a given small twist  (sufficiently small), the experienced reaction is given by M K    (1) This is just like the Hooke’s law for the springs. If a mass with moment of inertia I is attached to the rod, the torque will give the mass an angular acceleration  according to 2 2 d M I I dt      Then we get the following equation: 2 2 d θ K dt I    (2) Hence on solving this second order differential equation we get:      A t cos( ) (3)

T=2丌 Where, is the angular velocity of simple harmonic oscillation, and T is the period. So, if period Tand the K are known, the inertia of the rotated object can be expressed as K 4 The additive property of inertia: The moment of inertia of a composite system is the sum of the moments of inertia of its component subsystems(all taken about the same axis) =∑l(=123 Experiment device LED late spindle chopping bardetector aIrscrew sprIn power display ase 5 cycles reset Fig 2 Schematic diagram of the experiment device Procedure Familiarize yourself with the operation of the device. Adjust the device carefully so that it is ready for the measurement(i.e. the base is placed horizontally Determine the torsional constant K of the spiral spring with a plastic cylinder. Employ the theoretically calculated moment of inertia of the plastic cylinder as a known value Measure the moment of inertia of differently shaped objects. Compare the measured results with the theoretically calculated values Optional Verify the parallel axis theore

8 K I   (4) 2 T I K   (5) Where,  is the angular velocity of simple harmonic oscillation, and T is the period. So, if the period T and the K are known, the inertia of the rotated object can be expressed as: 2 2 4 K I T   (6) The additive property of inertia: The moment of inertia of a composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis): ( 1,2,3...) j j I I j    (7) Experiment device Fig. 2 Schematic diagram of the experiment device Procedure  Familiarize yourself with the operation of the device. Adjust the device carefully so that it is ready for the measurement (i.e. the base is placed horizontally.).  Determine the torsional constant K of the spiral spring with a plastic cylinder. Employ the theoretically calculated moment of inertia of the plastic cylinder as a known value.  Measure the moment of inertia of differently shaped objects. Compare the measured results with the theoretically calculated values. Optional  Verify the parallel axis theorem

Questions 1. On which factors is the moment of inertia dependent 2. What are the causes that may bring error to our measurement? References 1.沈元华,陆申龙基础物理实验( Fundamental Physics Laboratory)高等教育出版社北 京2004pp.100-103 2.RavitejUppuTorsionalPendulumhttp://www.cmi.ac.in/-ravitej/lab/4-torpen.pdf Appendix: l1 Uncertainty of Plastic cylinder: D&u(Ipr) u(m) (D) D u(D)=√l2(D)+n2(D) l(m)=√x2(m)+2(m Where, uBI(m)=d=0.1g u(D)= ∑(D-D 0.2 n(n-1) g 0.02 [2 Uncertainty of K l(K)= l(c)|a(72-7) 70 Please note: (2-73)=√(2)3+[(7)&7)=27×(T) Since we measure the time of 5 periods, so for each period, we have T==t and ()=√:()+2( ∑ n(n-1) 0.01 √3√3 S

9 Questions 1. On which factors is the moment of inertia dependent? 2. What are the causes that may bring error to our measurement? References 1. 沈元华，陆申龙 基础物理实验 (Fundamental Physics Laboratory) 高等教育出版社 北 京 2004 pp. 100-103 2. Ravitej Uppu Torsional Pendulum http://www.cmi.ac.in/~ravitej/lab/4-torpen.pdf Appendix: [1] Uncertainty of Plastic cylinder: 1 1 2 2 2 8 PC I mr mD   & 2 2 ( ) (D) ( ) 2 PC PC u m u u I I m D                 Where, 1 2 2 2 1 2 ( ) ( ) ( ) ( ) 0.1 0.2 ( ) 3 3 B B B B u m u m u m u m d g a u m g       & 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( 1) 0.02 ( ) 3 3 A B i A B u D u D u D D D u D n n a u D mm         . [2] Uncertainty of K. 2 2 2 2 0 2 2 0 ( ) ( ) ( ) PC PC u I u T T u K K I T T                  . Please note: 2 2 2 2 2 2 0 0 u T T u T u T ( ) [ ( )] [ ( )]    & 2 u T T u T ( ) 2 ( )   . Since we measure the time of 5 periods, so for each period, we have 1 5 T t  and 1 ( ) ( ) 5 u T u t  . Where, 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( 1) 0.01 ( ) 3 3 A B i A B u t u t u t t t u t n n a u t s        