PhysicsDepartment - Brock · PDF fileThe pendulum apparatus consists of a vertical post and ... lab report. ☛ ! At the ... The transmitted signal propagates in a narrow conical beam

Brock University Physics DepartmentSt. Catharines, Ontario, Canada L2S 3A1

PHYS 1P91 Laboratory Manual

Physics Department

Copyright c© Brock University, 2015-2016

Contents

1 The pendulum 1

2 Angular Motion 10

3 Collisions and conservation laws 18

4 The ballistic pendulum 26

5 Harmonic motion 34

A Review of math basics 42

B Error propagation rules 45

C Graphing techniques 47

i

ii

(ta initials)

first name (print) last name (print) brock id (ab17cd) (lab date)

Experiment 1

The pendulum

In this Experiment you will learn

• the relationship between the period and length of an object swinging in a gravitiational field;

• how approximate methods are used to simplify the analysis of a physical system;

• how to set and check the calibration of a data gathering device;

• to extend your data analysis capabilities with a computer-based fitting program;

• to apply different methods of error analysis to experimental results.

Prelab preparation

Print a copy of this Experiment to bring to your scheduled lab session. The data, observations andnotes entered on these pages will be needed when you write your lab report and as reference materialduring your final exam. Compile these printouts to create a lab book for the course.

To perform this Experiment and the Webwork Prelab Test successfully you need to be familiar with thecontent of this document and that of the following FLAP modules (www.physics.brocku.ca/PPLATO).Begin by trying the fast-track quiz to gauge your understanding of the topic and if necessary review themodule in depth, then try the exit test. Check off the box when a module is completed.

FLAP PHYS 1-1: Introducing measurement

FLAP PHYS 1-2: Errors and uncertainty

FLAP PHYS 1-3: Graphs and measurements

FLAP MATH 1-1: Arithmetic and algebra

FLAP MATH 1-2: Numbers, units and physical quantities

WEBWORK: the Prelab Pendulum Test must be completed before the lab session☛✡

✟✠! Important! Bring a printout of your Webwork test results and your lab schedule for review by theTAs before the lab session begins. You will not be allowed to perform this Experiment unless therequired Webwork module has been completed and you are scheduled to perform the lab on that day.

☛✡

✟✠! Important! Be sure to have every page of this printout signed by a TA before you leave at the endof the lab session. All your work needs to be kept for review by the instructor, if so requested.

CONGRATULATIONS! YOU ARE NOW READY TO PROCEED WITH THE EXPERIMENT!

1

2 EXPERIMENT 1. THE PENDULUM

Motion of the plane Pendulum

A simple pendulum consists of a compact mass m suspended from a fixed point by a massless string oflength L as shown in Fig. 1.1. The mass is subjected to a gravitational force ~F = m~g.

Figure 1.1: Two-dimensional (plane) trajectory of the simple Pendulum

The equilibrium position of m is O. Here, the tension ~P exerted by the string on m is exactly equaland opposite to the weight m~g of the mass. However, if m is at an angle θ (in radians) with respect toO, then the weight will have a component tangent to the arc with magnitude FT = |m~g sin θ|. This forceacting on the pendulum determines the motion of the pendulum bob.

An equation of motion describes the behaviour of a system in terms of motion as a function of time.In our case, this equation determines the position S of the pendulum bob along the arc of swing at anytime t. This equation of motion cannot be solved exactly for a system acted on by a force FT = |m~g sin θ|.

Small angle approximation

By introducing some restrictions, the pendulum system can be modeled after a system that does have awell known equation of motion. If we require that θ ≪ 1 radians, then sin θ ≈ θ and the force can beapproximated by FT = −mgθ. This restoring force that is now proportional to θ and always pulls the massm towards O, obeys Hooke’s Law:

Fh = −kS

A mass m subjected to a force that obeys Hooke’s Law will oscillate with an amplitude A about anequilibrium point O at a rate ω in radians/s determined by the spring constant k in Newtons/m. Theequation of motion for the Hooke’s Law system can be solved exactly; it is a sinusoidal function

S = Asin (ωt) = Asin

(

2π

Tt

)

→ S = Asin

(

√

k

mt

)

(1.1)

with a period of oscillation T = 2π√

m/k. With FT = Fh, this same equation also describes the path ofthe pendulum bob along the arc S = Lθ. To derive a relationship for T in terms of L and g,

mgθ = kS = kLθ → k =mg

L, ∴ T = 2π

√

L

g(1.2)

This equation predicts that the period T is independent of the mass m of the bob and that T is alsoindependent of the angle θ through which the ball swings, as long as the approximation sin θ ≈ θ is valid.

3

For θ = 15◦ ≈ 0.2618 radians, there is a difference of approximately 1.2% between θ and sin θ. In practice,the period T increases with the amplitude of the swing: by 0.4% when θ = 15◦, 1.7% when θ = 30◦.

Procedure

In this experiment, you are going to explore the realtionship between the period T and length L of aswinging pendulum.

The pendulum apparatus consists of a vertical post and fixed arm from which the pendulum, an alu-minum ball of radius r, diameter d, is suspended by a light nylon string of negligible mass (m ≈ 0).

A sliding arm is used to adjust the pendulum swing length. If the length is properly adjusted, or cali-brated to the scale on the pendulum post, then the scale can be used to directly measure, or set, this length.

When calibrated, the scale displays the length of the string s from the bottom of the sliding arm,the pivot point of the pendulum, to the top of the ball, to a precision of one millimetre (mm).

This length s is not exactly equal to the pendulum length L since the ball is not a point-mass (r 6= 0);L is the sum of the length of the string s and half the ball’s diameter d given in Table 1.1:

L = s+ 0.5 d. (1.3)

To calibrate the pendulum:

1. Release the string to set the ball on the table.

2. Align the bottom of the sliding arm, labeled Index, with the zero mark on the scale. Adjust thestring length so that the top of the ball lightly contacts the bottom of the arm, ensuring that thestring is not stretched. Secure the string with the clamping nut.

To check the calibration:

1. Raise the arm away from the ball and carefully reposition the arm until it once again just contactsthe top of the ball.

2. The index at the bottom of the sliding arm should be at the zero mark on the scale. If it is not,repeat the calibration adjustments until the bottom of the arm is in line with the zero mark of thescale and the arm lightly contacts the top of the pendulum ball.

☛✡

✟✠! All members of the group must independently check the calibration and verify that the pendulumis properly setup. Before proceeding any further, have a TA review and grade your effort.

Data gathering and analysis using Physicalab☛✡

✟✠! At the start of every lab session, click on the desktop icon to open a new Physicalab application,then enter one or two Brock email addresses for the member(s) of the group. Without valid emails,you will not be able to send yourself the graphs made during the experiment for inclusion in yourlab report.

☛✡

✟✠! At the end of the lab session, be sure to close the Physicalab application, otherwise your email addresswill be accessible to the next group using the workstation.


You are going to use a computer-controlled rangefinder and the Physicalab software to monitor the changein distance over time of the pendulum bob from the device.

The rangefinder determines the distance to an object by transmitting an untrasonic ping and measuringthe time required to receive the echo of the original signal after it is reflected from the object.

The transmitted signal propagates in a narrow conical beam and as it is the echo from the nearestobject that is accepted, you will need to take care that you are pinging the pendulum bob and not thependulum post or some other structure in the cone of the beam.

As the pendulum bob swings toward and away from the rangefinder, the variation in distance y overtime x can be approximated by a sine wave. By fitting your data to the equation of a sine wave andanalysing the properties of this sine wave, you can determine the period T of the pendulum.

1. Mount the larger ball m1 and calibrate the pendulum. Adjust the sliding arm so that the stringlength s is approximately 0.3 m. Record the actual length to a precision of 1 mm (0.001 m).

2. Set the pendulum swinging in a straight line, keeping θ less than approximately 15◦ . Wait severalseconds to allow for any stray oscillations present in the bob to dissipate.

? Does the angle of swing need to be precisely 15◦? How would other choices of angle affect the results?

3. Shift focus to the Physicalab software. Check the Dig1 box and choose to collect 50 points at 0.1

s/point. Click Get data to acquire a data set of distance y as a function of time x.

4. Click Draw to generate a graph of your data. Your points should display a nice smooth sinewave, without peaks, stray points, or flat spots. If any of these are noted, adjust the position ofthe rangefinder and acquire a new data set. Flat spots occur when an object is too close to therangefinder.

? If the pendulum was not swinging in-line with but at a significant angle β to the rangefinder, howwould the appearance of your graph change? Try to explain what is going on mathematically. Wouldthis likely affect your results?

5. Select fit to: y= and enter A*sin(B*x+C)+D in the fitting equation box.

As reviewed in the Appendix, A is the amplitude of the sine wave, C is the initial phase angle (inradians) when x=0, and D is the average distance of the wave from the detector, i.e. when A=0.

The fit parameter B (in radians/s) is equivalent to ω, the rate of change in angle with time, so thatB*x is an angle in radians. If this angle value is set to 2π radians, then x=T , the period of the sinewave in seconds.

6. Click Draw . If you get an error message the initial guesses for the fitting parameters may be toodistant from the required values for the fitting program to properly converge.

Look at your graph and enter some reasonable approximate values for the fitting parameters. Youcan get an initial guess for B by estimating the time x between two adjacent minima, or one period,of the sine wave.

7. Label the axes and and give your graph a descriptive title that includes the string length s. Click

Send to: to email yourself and your partner a copy of the graph for later inclusion in your labreports. Note: do this for all the graphs created during the lab session.

5

8. Record in Table 1.1 the trial length s and the computer calculated value of the fitting parameter B(radians/second). Do a quick calculation for g to make sure that the value is reasonable.

? Why are you fitting an equation to your data when you can estimate the period T of the pendulumdirectly from the graph?

• Repeat the above steps for m1 with s = 0.45, 0.60, 0.75 and 0.90 m.

? Do you have to use these specific lengths? Could another set of values be used just as well? Howmight your choice of lengths affect the results?

• Mount the second ball m2, recalibrate the pendulum and verify the calibration, set the stringlength to approximately s = 0.5 m, and record this one measurement in Table 1.1.

Run, i mass m (kg) d (m) s (m) L (m) B (rad/s) T (s) T 2 (s2) gi (m/s2)

1 m1 0.0225 0.02540

2 m1 0.0225 0.02540

3 m1 0.0225 0.02540

4 m1 0.0225 0.02540

5 m1 0.0225 0.02540

1 m2 0.0095 0.01904

Table 1.1: Table of experimental results

Part 1: Determining g from a single measurement

Since you made only one trial using the small ball of mass m2, error propagation rules need to be usedto determine error estimates for L and g. The measurement errors in s, d, represented by ∆s and ∆d,are determined from the scales of the measuring instruments. The micrometer used to measure the balldiameter d had a resolution, or scale increment, of 0.00001 m.

∆s = ±............... ∆d = ±...............

• Equation 1.3 is used to calculate L and derive ∆L. Show in three steps below the relevant equation,the variables replaced by the appropriate unrounded values and finally show the numeric result. Unitsare not required at this step.

L = s+ 0.5d = ....................... = .......................

∆L =√

(∆s)2 + (0.5∆d)2 = ....................... = .......................

Present the final result for L±∆L, properly rounded and with the correct units

L = .............± .............


• Equation 1.2 expresses the relationship between g, L and T . You now have ∆L but not the error inT , ∆T . Since the value of T is derived from the fit parameter B and ∆B is also given by the fit,you could derive ∆T from ∆B. However, a more direct approach is to simply rewrite Equation 1.2in terms of B instead of T . Having done this, calculate g and the error ∆g in g.

g = ....................... = ....................... = .......................

∆g = ....................... = ....................... = .......................

g = .............± .............

Part 2: Determining g from a series of measurements

You performed five trials (i = 5) using the large ball of massm1 to obtain five results for g that are expectedto have the same value if Equation 1.2 is valid. In this case, you can invoke the theory of statistics toevaluate a sample average 〈g〉 for the five trials and a measure of the scattering of the trials around theaverage, or standard deviation, of the sample σ(g).

• Use Table 1.2 to calculate the average and standard deviation of the sample of g values, then sum-marize the result in the proper format below:

i gi ∆gi = gi − 〈g〉 (∆gi)2

1

2

3

4

5

〈g〉 = variance =

σ(g) =

Table 1.2: Calculation template for 〈g〉 and σ(g). Variance = σ(g)2 = 1N−1

∑N1 (∆gi)

2

g = .............± .............

• In Physicalab, enter in a column your five g values, then in the Options menu select Insert XIndex to add a column of index values. Select bellcurve to view your data as a distribution andcompare the mean and standard deviation values from the graph with your results from above. Theyshould be the same. Send a copy of the graph.

Hint: to view a more representative distribution with a large number of samples, click Options andload distribution. See how the graph changes as you vary the number of bins that the data ispartitioned into. Checking the bargraph box will display your data as a frequency bargraph.

7

Part 3: Determining g from the slope of a graph

A graphical method can also be used to determine g and ∆g. Here, a curve that represents the properrelationship between the x and y coordinates is drawn through your data so that the total distance fromthe data points to the line is a minimum. Rewriting Equation 1.2 in terms of L as a function of T 2

L =( g

4π2

)

T 2 (1.4)

yields a linear relationship y = mx+ b between y = L and x = T 2, with slope m = g/(4π2) and y-interceptb = 0.

• Enter the five coordinates (T 2, L) in the Physicalab data window. Select scatter plot. Click Drawto generate a graph of your data. Select fit to: y= and enter A*x+B in the fitting equation box.

Click Draw . The computer will evaluate a line of best fit through the data points and output thefit results. The χ2 value is a measure of the goodness of the fit; χ2 gets smaller as the total distanceof the data points from the curve of best fit decreases.

• Send the graph, then record below the slope A and error ∆A

A = .............± .............

• Calculate values for g and ∆g.

g = ....................... = ....................... = .......................

∆g = ....................... = ....................... = .......................

g = .............± .............

• Redraw the preceding graph, this time including the data point from the single trial using mass m2

and repeat the liinear fit. This graph can provide you with insight on the dependence of g on themass of the pendulum bob. Record the slope for comparison with the previous result.

A = .............± .............

☛✡

✟✠! Before you leave, have a TA enter your five g values into the pendulum database. You can thensee how your results compare with the group average of all the students who have already done theexperiment.

As the data from all the different groups of students doing the experiment is accumulated, a nicestatistical distribution of the value of g should evolve and the standard deviation σ(g) should sys-tematically decrease.

Lab report

Go to your course homepage on Sakai (Resources, Lab templates) to access the online lab report worksheetfor this experiment. The worksheet has to be completed as instructed and sent to Turnitin before the lab


report submission deadline, at 11:00pm six days following your scheduled lab session. Turnitin will notaccept submissions after the due date. Unsubmitted lab reports are assigned a grade of zero.

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Check your schedule!

This is a reminder that Experiment 2 does not necessarily follow Experiment 1.

You need to login to your course homepage on Sakai and check your lab schedule to determine theexperiment rotation that you are to follow.

My Lab dates: Exp.2:......... Exp.3:......... Exp.4:......... Exp.5.........

Note: The Lab Instructor will verify that you are attending on the correct date and have prepared for thescheduled Experiment; if the lab date or Experiment number do not match your schedule, or the reviewquestions are not completed, you will be required to leave the lab and you will miss the opportunity toperform the experiment. This could result in a grade of Zero for the missed Experiment.

To summarize:

• There are five Experiments to be performed during this course, Experiment 1, 2, 3, 4, 5.

• Everyone does the first experiment on their first scheduled lab session.

• The next four experiments are scheduled concurrently on any given lab date.

• To distribute the students evenly among the scheduled experiments, each student is assigned to oneof four groups, by the Physics Department. Your schedule is shown in your course homepage onSakai.

9

(ta initials)


Experiment 2

Angular Motion

A friendly reminder: Is this your scheduled experiment? See the previous page.


• some basic principles of rotational motion around a central pivot point

• to measure the specific rotational inertia Id of a solid disc by a multistep method

• to manipulate equations and have them disclose information of interest to you

• to use a computer-based fitting program to enhance the analysis of your data

• to apply error analysis to experimental results and thus make your results relevant.

Prelab preparation



FLAP PHYS 1-1: Introducing measurement


FLAP MATH 1-4: Solving equations

FLAP MATH 1-6: Trigonometric functions

Webwork: the Prelab Angular Motion Test must be completed before the lab session☛✡


☛✡



10

11

Torque acting on a rotating body

To understand what torque is, let us consider a rigid body with its centre of mass at a point r = 0,constrained to rotate about that point. The radius vector ~r of a point on the body has the origin at r = 0and ends at the point, a distance r from the origin, as shown in Figure 2.1.

Figure 2.1: Rotational coordinates

If a force ~F is applied at that point, in the planeof rotation of the disc, at a distance r from thecentre, and at an angle φ to the radius vector ~r,the magnitude of the turning torque τ produced bythis force is:

τ = rF sinφ (2.1)

The torque τ depends both on the distance fromthe centre of rotation and on the direction of theapplied force ~F . Decomposing ~F into a radial com-ponent ~Fr and a tangential component ~Ft so that~F = ~Fr + ~Ft, it can be seen that only the tan-gential component of ~F causes torque and affectsthe magnitude of τ . A force applied through thecentre of rotation has a zero tangential component(φ = 0, Ft = 0) and the radial component aloneproduces no torque and will not cause angular ac-celeration. As the angle of the force changes, sodoes the torque experienced by the body; for a given magnitude of the force, F , the maximum torque isproduced when ~F = ~Ft is perpendicular to ~r and τ = rFt = rF .

Resisting the force ~F is the total mass M of the rotating body. Suppose that this mass consists ofmany particles of mass mi so that M =

∑

imi. In a translational motion, the force acts equally on all the

component particles of the body at once, according to Newton’s second Law ~F =∑

i(mi~a) = M~a. In a

rotation, the rotational effect of ~F on a particle is proportional to the distance r from the centre of rotation.While the entire rigid body experiences a single angular accelaration α, common to all its particles, eachof them will experience a tangential acceleration ai that is proportional to the distance ri from the centreof rotation.

The rotational moment of inertia I of a rigid body composed of many particles is simply the sum ofthe individual rotational moments of inertia of all particles:

I =∑

i

mir2i (2.2)

For a thin hoop, where all the particles are located at the same common radius away from the centre,ri = R, Eq. 2.2 reduces to I = MR2; for other shapes, the calculation may not be so simple.

The rotational form of the Newton’s Second Law is τ = Iα:

F =∑

i

(mia) = Ma −→ τ =∑

i

(mir2i α) = Iα.

The torque τ plays the same role for rotational motion that the force F plays for translational motion.

Determining the rotational inertia of a disc

Consider a homogeneous disc of radius R and mass M constrained to rotate without friction around thecentre of mass. A massless string is wrapped around the outer edge of the disc and connected to a massm that is subjected to the force of gravity Fg, as shown in Figure 2.2. The string experiences a tension T

12 EXPERIMENT 2. ANGULAR MOTION

due to the weight of m; at the other end of the string the same tension T acts on the edge of the disc at adistance r from the centre of rotation. The displacement of the falling mass is given by Equation 2.3:

Figure 2.2: A falling block causes the disc to rotate

y = y0 + v0t+ at2/2 (2.3)

When a force is applied, the disc, initially at rest,begins to spin as m falls with linear acceleration

a = g − T

m. (2.4)

The rotational acceleration of the disk is

α =a

r=

τ

I=

Tr

I. (2.5)

By combining Eqs. 2.4 and 2.5, the rotational iner-tia I of the disc can be expressed in terms of a asfollows:

I = mr2(g

a− 1)

. (2.6)

Measuring the acceleration of the falling mass, andcomparing it to the acceleration of the free fall,yields an operational measurement of the momentof inertia of the disk.

Procedure

The rotational inertia Id of a steel disc will be determined indirectly in three steps:

1. measure the rotational inertia Ic of a spinning cradle that will accept the disc;

2. measure the rotational inertia It = Ic + Id of the cradle and disc combination;

3. calculate Id from the difference of these two results.

Figure 2.3: Experimental setup

The cradle consists of a plastic disc boltedto a metal drum of four stacked pulleys ofvarying diameters. The cradle is centeredon a ball-bearing post and can rotate nearlyfriction-free around this vertical axis as longas it is not rubbing against the side of thepost.

A string, wrapped around one of the drum pul-leys and attached at the other end to a suspendedmass m, allows a torque τ to be applied to the cra-dle at radius r of the pulley. The string leaving thedrum must be perpendicular to the axis of rotationand to the radius vector, as shown in Figure 2.3.

? Suppose that the string was not perpendicu-lar. What adjustments, if any, would you need to make to your torque calculations?

13

The pulley A in Figure 2.3 is part of a photogate system. The C-shaped photogate has an infrared trans-mitter at one end and an infrared detector at the other end. As the pulley rotates, the spokes interruptthe beam, turning on and off the red light-emitting diode (LED). The LabPro device detects and countsthese changes and Physicalab calculates the distance that the string has moved.

The pulley has ten spokes and a circumference of 155 mm, so there are twenty pulses transmitted everyrotation of the pulley. Each pulse represents a radial distance of 7.75 mm travelled by the string and hencethe same distance y moved by the mass m.

After the data acquisition is initiated, LabPro waits for the first transition and marks this event withthe elapsed time and assigns it an initial distance y = 0. After all subsequent transitions, the new timeand total distance travelled are sent. Note that the time t = 0 set by LabPro will not likely coincide withthe start of the motion.

Data gathering and analysis using Physicalab

1. Wrap the string, trying not to overlap the strands, around the second smallest pulley (r = 0.02282 m)on the cradle and arrange the string path as shown in Figure 2.3.

2. Place a weight mw on the mass holder and rotate the cradle to raise these to the top of the assembly.Hold the cradle in place by putting a weight on the edge of the platform.

? Consider the geometry of the system, is the cradle likely to rub against the post?

3. Shift focus to the Physicalab software. Select Dig2, the channel that the photogate should beconnected to, then choose to collect a nominal 20 points with 0.5 seconds between points.

4. Remove the weight from the platform. The cradle will begin to rotate. Press Get data to startgathering the photogate data. Note that the platform will not likely have a zero velocity v0 = 0, attime t = 0 when the data acquisition begins.

? Will this arbitrary starting point in your data collection have any effect on the quantity that youwant to determine from the fit?

5. Stop the data acquisition and the platform rotation before the mass holder reaches the end of it’s

trajectory; it will save you the trouble of having to re-spool the string on the pulleys.

6. At the end of the run, review your graphed data; it should approximate a smooth curve that representsthe falling of a mass m under constant acceleration a. If this is not the case, repeat the trial.

? Could you delete a portion of your data set and still get the desired result from your fit? How mightthe result be affected by such a change?

7. Select fit to: y= and enter A+B*x+C*x**2/2 in the fitting equation box. Click Draw to

perform a fit of your data. Click Send to: to email yourself and your partner a copy of the graphfor later inclusion in your lab reports.

? How do you determine which of the fit parameters represents the acceleration a?


mw (kg) 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045

a1 (m/s2)

a2 (m/s2)

〈a〉 (m/s2)

Table 2.1: Acceleration data for unloaded cradle assembly

Part 1: Determining the rotational inertia of the cradle

• Fill in Table 2.1, then calculate an average acceleration 〈a〉 for the various masses.

You may now wish to evaluate the rotational inertia Ic for the unloaded cradle, using Equation 2.6 andthe value of 〈a〉 for a given mw in Table reftab:cradledata.

This would not likely lead to the correct Ic value because m in Equation 2.6 represents the total massm = mw +mh −mf that causes the tension T on the string. In your experiment, the effect of the appliedmass mw and that of the mass holder mh is diminished by an unknown mass mf required to overcome thestatic and dynamic friction experienced by the platform and pulleys.

You will have noted that the cradle sometimes begins to rotate with only the mass holder attached,when mw = 0. In this case, mh > mf and to prevent the system from rotating, mw would need to benegative (or the mass holder would need to be lighter).

To get around the difficulty of not knowing mf , you could determine mh by weighing the mass holder,then use two sets of values from Table 2.1 and solve two simultaneous equations to eliminate the unknownvariable mf .

A better method makes use of the whole set of tabulated data values (〈a〉,mw) that are available whilemaking unnecessary a knowledge of both mh and mf .

Begin by rearranging Equation 2.6 so that mw is expressed as a function of a:

Ic = (mw +mh −mf )r2(g

a− 1)

≈ (mw +mh −mf )r2(g

a

)

→ mw =

(

Icgr2

)

a+ (mf −mh). (2.7)

The simplification is valid only if a ≪ g so that (g/a− 1) ≈ (g/a). Then the resulting equation is thatof a straight line with slope Ic/(gr

2) and y-intercept (mf −mh).

• Enter into an empty data window your set of (〈a〉,m) coordinates. The scatter plot of your datashould show a linear behaviour. Select fit to: y= and enter A+B*x in the fitting equation box.

Click Draw and record below the fit parameters A and B along with their appropriate units.

A = ...............± ............... B = ...............± ...............

• Calculate Ic using Equation 2.7 and use the appropriate error propagation rules to evaluate thecorresponding uncertainty ∆Ic:

Ic = ....................... = ....................... = .......................

15

∆Ic = ....................... = ....................... = .......................

Ic = .............± .............

Part 2: Determining the rotational inertia of cradle and disc

• Use the peg on the steel disc to center it on the cradle, then place a small piece of paper under thedisc to prevent it from slipping. Repeat the previous steps to determine the rotational inertia It ofcradle-plus-disc.

mw (kg) 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100

a1 (m/s2)

a2 (m/s2)

〈a〉 (m/s2)

Table 2.2: Acceleration data for cradle assembly with disc

A = ...............± ............... B = ...............± ...............

It = ....................... = ....................... = .......................

∆It = ....................... = ....................... = .......................

It = .................± .................

Part 3: Determining the rotational inertia of the disc

• Subtract Ic from It to obtain Id, the rotational inertia of the disc alone.

Id = ....................... = ....................... = .......................

∆Id = ....................... = ....................... = .......................

Id = .................± .................

• Measure the mass M and radius R of the steel disc. Also weigh the mass mh of the mass holder.

M = .........± ......... R = .........± ......... mh = .........± .........


• Use the appropriate equation to calculate the theoretical rotational inertia for the disc that you used,as well an estimate of the error.

Id(th) = ...................... = ...................... = ......................

∆Id(th) = ...................... = ...................... = ......................

Id(th) = .................± .................

• Perform a trial using the smallest pulley with pull mass m = 30 g on the unloaded cradle.Compare this a with the previous result for a obtained using the second smallest pulley and the samepull mass. The pulley radii from smallest to largest are: 1.539 cm, 2.282 cm, 3.350 cm and 4.343 cm.

a(r = 1.539) = ................., a(r = 2.282) = .................

? Do these two a values obtained by varying the pulley radius r agree with the results predicted bythe theory?

Lab report

Go to your course homepage on Sakai (Resources, Lab templates) to access the online lab report worksheetfor this experiment. The worksheet has to be completed as instructed and sent to Turnitin before the labreport submission deadline, at 11:00pm six days following your scheduled lab session. Turnitin will notaccept submissions after the due date. Unsubmitted lab reports are assigned a grade of zero.

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Experiment 3

Collisions and conservation laws


• about the principle of the conservation of momentum

• the law of conservation of energy applied toQ, a ration that measures energy loss during an interaction

• to make a collision record of two colliding pucks

• to analyse the quality of vector dot data and properly draw vectors on a sheet of paper

• to correctly perform a graphical vector addition, with careful transposition and measurement ofvectors

• that you can apply a divide-and-conquer self-checking strategy to complex expressions, so that thechances of algebraic or calculation errors are greatly reduced.

Prelab preparation




FLAP MATH 2-1: Introducing geometry

FLAP MATH 2-4: Introducing scalars and vectors

Webwork: the Prelab Colliding Pucks Test must be completed before the lab session☛✡


☛✡



18

19

Conservation of momentum

The velocity ~v of a body of mass m is determined by its speed, a scalar quantity, and its direction of motion,represented by a unit vector. If we consider a collision of two such masses m1 and m2 with velocities ~v1b and~v2b before the collision, and velocities ~v1a and ~v2a after the collision, we note that it is generally difficult,if not impossible, to predict these resulting velocities ~v1a and ~v2a. To accomplish this, one would need tohave a complete knowledge of the physical characteristics of the objects (size, shape, et cetera) and of thegeometry of the interaction.

The linear momentum ~p of a body is a vector equal to the product of its mass m (a scalar) and itsvelocity ~v (a vector). The law of conservation of linear momentum states that:

If there are no net external forces acting on the masses, then the total momentum ~pb before the

collision is equal to the total momentum ~pa after the collision.

A mathematical formulation of this law for a collision between two masses m1 and m2 is a vector equationand can be expressed as follows:

~p1b + ~p2b = ~p1a + ~p2a

m1~v1b +m2~v2b = m1~v1a +m2~v2a. (3.1)

Rearranging Equation 3.1 in terms of the change in the momentum ∆~p1 of m1 and ∆~p2 of m2 revealsthat the net change will be zero and that the vectors will be oriented anti-parallel to one another:

~p1a − ~p1b + ~p2a − ~p2b = 0 → (~p1a − ~p1b) = −(~p2a − ~p2b) → ∆~p1 = −∆~p2

Similarly, the change in the velocity for the two masses will result in two anti-parallel vectors:

m1(~v1a − ~v1b) = −m2(~v2a − ~v2b) → m1∆~v1 = −m2∆~v2 (3.2)

In particular, for equal masses m1 = m2, the magnitudes of changes in the velocity of the two massesshould be the same.

Since the two vectors ∆~v1, ∆~v2 are colinear, the vector Equation 3.2 can be reduced to a scalar equationand expressed in terms of the magnitudes of the vector changes in velocities |~v2a − ~v2b| and |~v1a − ~v1b|.Rearranging Equation 3.2 as a ratio of masses m1 and m2:

m1

m2=

|∆~v2||∆~v1|

=|~v2a − ~v2b||~v1a − ~v1b|

. (3.3)

It is also of interest to know whether kinetic energy K = mv2/2 is conserved during a collision. The“quality factor” Q = Ka/Kb is defined as the ratio of the total kinetic energy Ka after the collision to thetotal kinetic energy Kb before. The kinetic energy, and therefore Q, are scalar quantities. In an elastic

collision, the kinetic energy of the system is conserved, and so Q = 1; in an inelastic collision some ofthe kinetic energy may be transformed into heat and sound energy during the collision, or there could befrictional forces acting on the masses during the interaction, and so Q < 1. Note that Q depends on thesquare of the velocity and hence will be very sensitive to variations in v.

Q =Ka

Kb=

12 m1 v

21a +

12 m2 v

22a

12 m1 v

21b +

12 m2 v

22b

=m1 v

21a +m2 v

22a

m1 v21b +m2 v22b(3.4)

In principle, in a superelastic collision, the total translational kinetic energy may even increase (Q > 1),for example if some rotational kinetic energy imparted onto the pucks before the collision transfers intotranslational kinetic energy, or if additional energy is released by the collision itself (e.g., a collision of twospring-loaded mousetraps).

20 EXPERIMENT 3. COLLISIONS AND CONSERVATION LAWS

Procedure

In this experiment the mass ratio of two colliding pucks will be calculated to determine whether or notlinear momentum was conserved during the collision. The ratio Q will be used to estimate the kineticenergy lost during the interaction.

Figure 3.1: Trails left by moving pucks

The equipment consists of a flat glass plate in a metalframe which can be levelled by varying the height of fouradjustable legs. The glass plate is covered by a conduc-tive black pad. A sheet of regular white paper is placedover this conducting layer. The sheet must be flat andfree of any kinks or other deposits.

The collision is between two heavy metal pucks. Eachpuck is tethered to a plastic hose that provides the puckwith compressed air and carries within it an electrodewire. The hose should hang freely, without any twists orinterference from the rest of the equipment.

When turned on, the air exits from a hole at the bot-tom of the puck, creating a high pressure layer betweenthe puck and the paper, causing the puck to levitate and move with negligible friction. When the frame isproperly levelled, the puck will float in place, without any lateral movement.

A spark timer is used to provide the high voltage pulses required in the experiment. One terminal ofthe timer is connected through the electrode wire in the hose to an insulated needle at the bottom of thepuck.

The other terminal of the spark timer is connected to the conductive sheet, completing the electriccircuit. Sparks are generated from the needle through the white paper to the conducting paper beneath.These pulses produce black spots on the bottom of the white paper, marking the positions of the pucks atequal time intervals u. Puck speeds would then be expressed in units of cm/u or mm/u.

The firing rate in units of u−1 of the timer can be adjusted with a control knob on the unit. The timerwill produce sparks when the pushbutton switch at the front of the unit is pressed. To make a collisionrecord:

1. Place a new sheet of white paper over the conducting layer, making sure that the paper is flat.

2. Very slowly turn on the air until the pucks begin to float. Excessive air pressure will burst the airhose. Level the frame by adjusting the height of the four legs until the pucks float in place.

3. Place the pucks against the launchers in adjacent corners of the frame, and release them toward thecentre of the paper, where they will collide. Do several trial runs. While any collision is theoreticallyvalid, for ease of analysis your collision should be fairly symmetric as in Figure 3.1 and take about2-3 seconds to complete.

Small initial velocities will cause the pucks to slow down due to friction and yield incorrect results.The pucks should not rotate as this would give them rotational kinetic energy. We are concernedonly with translational kinetic energy.

4. To record the collision, instruct an assistant to press and hold the switch of the spark timer justbefore you release the pucks, and to release it when the pucks rebound from the edges of the frame.Every member of the team needs to make their own collision record to analyse.

5. The mass of each puck has been measured and is displayed on the puck. Assign this mass value tothe corresponding trace for future reference, keeping in mind that the traces appear on the bottom

21

of the paper. Be sure to match each mass with the corresponding trace, otherwise your Q valuewill not make sense.

6. A good collision record will show for each of the four trails a series of 4-6 equadistant and colineardots, spanning a length greater than 100 mm. The length is measured with a ruler scale graduatedin millimetres. If your collision did not turn out, flip the sheet of paper paper and try again.

? Is the suggested minimum length of 100 mm significant? Does the measurement error depend on thelength of the a vector? If so, how would you minimize the measurement error?

Data analysis

Figure 3.2: Join dots to obtain vectors

It can be seen from Equation 3.3 that the genera-tion of the vectors ~v1b, ~v2b, ~v1a, and ~v2a is required.For example, the vector ~v1b can be obtained fromthe collision record by joining a series of dots alongthe Puck 1 trail (before the collision) spanning ntime intervals.

• Draw four proper vectors on your sheet,(line segment begins and ends at a dot, arrow-head ends at a dot) as shown in Figure 3.2.It is important to use this same number n oftime intervals for all the vectors so that youare, in effect, applying the same time scalen ∗ u to each vector.

• Measure the distance between the adjacent dots of each trail and enter the results and observationsin Table 3.

vector |v1b| |v1a| |v2b| |v2a|

number of segments in vector

dot spacing first segment (mm)

dot spacing last segment (mm)

vector length (mm)

dots are colinear? yes/no yes/no yes/no yes/no

Table 3.1: Experimental data: vector analysis results

? Evaluate the quality of your collision record. Did the distance between the dots in each trail remainthe same and did these dots form a straight line? Is the trail a valid representation of a vector?


m1 |v1b| |v1a| |~v1b − ~v1a| Time units m2 |v2b| |v2a| |~v2a − ~v2b|(g) (mm) (mm) (mm) (u) (g) (mm) (mm) (mm)

Table 3.2: Experimental data: masses and vector magnitudes

Figure 3.3: Transposition of a vector

The vectors (~v2a − ~v2b) and (~v1b − ~v1a) dependon the magnitude and direction of the compo-nent vectors, hence a graphical vector additionmust be performed. A reasonably long vector or linesegment with well defined endpoints can be trans-posed very accurately by visually estimating the fi-nal placement of this vector as shown in Figure 3.3:

• Using a long ruler, extend the line segmentAB beyond the region of point D.

• Place the ruler so that the edge rests on pointC and is parallel to the previously drawn line.

• Draw a line through C beyond points A andD.

• Measure near A and D the perpendiculardistance between the two lines to verify thatthey are indeed parallel.

• Determine the length AB with a ruler anddraw a line segment of length AB from C toD to define the vector CD. Enter your resultsin Table 3.

☛✡

✟✠! When your vector transpositions are completed, have a TA review your collision record and vectoraddition. Make sure to have the TA sign and date your collision record. You need to keep thisexperimental data as part of your lab notes for possible use at a later date.

From Equation 3.2, the change in momentum of one puck is equal and opposite to that of the otherpuck. The two resultant vectors should then be nearly equal in magnitude since m1 ≈ m2 but now parallel:note how (~v1a − ~v1b) = −(~v1b − ~v1a) would change Equation 3.2.

? Compare the lengths of vectors (~v2a −~v2b) and (~v1b −~v1a). Does the result make sense in light of thepreceding explanation?

• To quantify the test for parallelism, extend the vector lengths as above and calculate the angle θ, indegrees, between the two extended line segments. It is given by θ = arctan(dy/dx), where dy = y2−y1is the difference in the separations y1 and y2 at the opposite ends of the two lines of length dx.

dx = ............. y1 = ............. y2 = .............

23

dy = ..................... = ..................... = .....................

θ = ..................... = ..................... = .....................

• Determine a theoretical value (m1/m2) and error ∆m1/m2 for the mass ratio from the given valuesof m1 and m2. The measurement error for these masses is ±0.1 g.

m1

m2= ..................... = ..................... = .....................

∆

(

m1

m2

)

= ..................... = ..................... = .....................

(Theoretical)m1

m2= ...............± ...............

• Use Equation 3.3 to calculate a value and error for the experimental mass ratio of the pucks.

m1

m2= ..................... = ..................... = .....................

∆

(

m1

m2

)

= ..................... = ..................... = .....................

(Experimental)m1

m2= ...............± ...............

• Use Equation 3.4 and the theoretical m1 and m2 values determine Q and ∆Q.

When solving equations such as for Q and ∆Q, a good technique is to split the equation into severalterms and solve for each term separately. This avoids repetition and will expose possible calculation errors.

For example, there are four identical mv2 terms, each of which appears several times in the Q and ∆Qequations. Evaluate each term only once, then compare the four results. You would expect them to besimilar since the m and v values are also similar.

Likewise, the equations for each error term are identical in form and should also yield similar results.You can then confidently complete the calculation of Q and ∆Q.

A = m1v21a = ..................... = .....................

B = m2v22a = ..................... = .....................

C = m1v21b = ..................... = .....................


D = m2v22b = ..................... = .....................

Q =A+B

C +D= ..................... = .....................

∆A = A

√

(

∆m1

m1

)2

+

(

2∆v1av1a

)2

= ..................... = .....................

∆B = B

√

(

∆m2

m2

)2

+

(

2∆v2av2a

)2

= ..................... = .....................

∆C = C

√

(

∆m1

m1

)2

+

(

2∆v1bv1b

)2

= ..................... = .....................

∆D = D

√

(

∆m2

m2

)2

+

(

2∆v2bv2b

)2

= ..................... = .....................

∆Q = Q

√

(∆A)2 + (∆B)2

(A+B)2+

(∆C)2 + (∆D)2

(C +D)2= ..................... = .....................

Q = ...............± ...............

Note: Keep your collision record for your own further analysis or for review by the TA or professor, ifrequested.

IMPORTANT: BEFORE LEAVING THE LAB, HAVE A T.A. INITIAL YOUR WORKBOOK!

Lab report

Go to your course homepage to access the online lab report worksheet for this experiment. The worksheethas to be completed as instructed and sent to Turnitin before the lab report submission deadline, at11:59pm six days following your scheduled lab session. Turnitin will not accept submissions after the duedate. Unsubmitted lab reports are assigned a grade of zero.

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Experiment 4

The ballistic pendulum


• how to determine the speed of a projectile as well as the energy of the launcher that fired it

• that real-world interactions typically involve a series of energy exchanges

• the energy relationship between the linear motion of a ball and the angular motion of a pendulum

• that successful experimental results rely on careful measurement and application of technique



Prelab preparation




FLAP MATH 1-1: Arithmetic and algebra

FLAP MATH 1-2: Numbers, units and physical quantities

Webwork: the Prelab Ballistic Pendulum Test must be completed before the lab session☛✡


☛✡

✟✠! Important! Be sure to have every page of this printoutsigned by a TA before you leave at the end ofthe lab session. All your work needs to be kept for review by the instructor, if so requested.


26

27

Figure 4.1: Ballistic Pendulum

In this experiment we explore the transfer andconservation of energy and momentum in a colli-sion of two objects. One of these objects is a smallprojectile of mass m that is given a certain velocityv by a launcher. The second object is a stationarypendulum. As the projectile hits the pendulum,a re-distribution of energy and momentum takesplace.

In a certain class of collisions, the projectile iscaptured by the pendulum. For such inelastic col-

lisions, there is only one combined object that ismoving at the end, and carries all of the kinetic en-ergy K and momentum P . If the mass of the pen-dulum is M then the total mass of the combinedobject at the end of the collision is MT = M +m.If the bob is stationary when the projectile hits, itcontributes nothing to the total kinetic energy and momentum of the system before the collision. Thusthe Law of Conservation of Momentum in this case yields:

Pbefore = mv = Pafter = (M +m)vT = MT vT (4.1)

where vT is now the velocity of the combined object immediately after the collision. Knowing v and MT ,we can use Equation 4.1 to determine vT and show that the kinetic energy is not conserved:

Kafter =1

2MT v

2T =

1

2MT

(

mv

MT

)2

=m

MT

(

mv2

2

)

=m

MTKbefore < Kbefore. (4.2)

As the pendulum begins to swing after the collision, another physical process takes place; a conversionof the kinetic energy of the moving object into into gravitational potential energy as it swings up, losingkinetic energy and gaining potential energy. At the bottom of the swing, the pendulum of mass MT hasall of its energy in the form of kinetic energy. At the top of the swing, all of the pendulum’s energy isconverted into gravitational potential energy, and as MT momentarily pauses and reverses its motion, thekinetic energy falls to zero. Thus we can write:

Ebottom = K =1

2MT v

2T = Etop = MT gh. (4.3)

Combining Equations 4.1 and 4.3 to eliminate vT gives us an expression that relates the initial velocity vof the projectile to the final elevation h of the combined object.

v =MT

m

√

2gh, (4.4)

The length Rcm describes the radius of the arc from the pivot point to the centre of mass of combined rod,block, and block contents. With the vertical orientation of Rcm as the base of a right-angled triangle, hcan be expressed in terms of the angle θ of the swing: R cos θ = (R− h). Equation 4.4 then becomes

v =MT

m

√

2gRcm (1− cos θ) (4.5)

There is another energy conversion taking place, even before the collision. In the experiment, thelauncher gives the projectile its initial kinetic energy and momentum by releasing the potential energy

28 EXPERIMENT 4. THE BALLISTIC PENDULUM

and momentum of a compressed spring an converting it into the kinetic energy of the moving projectile.Conservation of energy in this case yields

Einitial = PE =1

2kx2 = Efinal = K =

1

2mv2 (4.6)

where x is the spring displacement from its equilibrium (relaxed) length and k is the spring constant. Asthe projectile is released from rest, the projectile has no initial kinetic energy. As the projectile begins toaccelerate, a conversion of energy takes place where the potential energy stored in the spring is convertedinto the kinetic energy of the projectile. At the point where the projectile releases from the spring, all ofthe spring’s stored potential energy has been converted into the projectile’s kinetic energy. We can thendetermine a value for the spring constant k of the launcher by equating the initial potential energy of thespring to the final kinetic energy of the projectile:

k =mv2

x2(4.7)

Procedure

The launcher component of the ballistic pendulum consists of a precision spring encased in an aluminumbarrel. One end of the spring is secured to the closed end of the barrel. The other end is attached to apiston that slides along the inside of the barrel.

The projectile rests on the face of the piston. A trigger mechanism allows the piston to be locked inone of three force settings: short, medium or long range.

When the launcher is in the discharged position, the spring is subjected to a small compression so thatit will not rattle when released. This preload x0 is in the order of 1 mm. For this experiment, we willassume that when the launcher is discharged the potential energy stored in the spring is approximately zero.

The pendulum consists of a rod and bob of combined mass M attached to a pivot point. When animpact takes place, the pendulum catches the impacting mass m, changing the total mass of the pendulumto MT = M +m, and is caused to swing about the pivot point to a maximum angle of deviation θ, relativeto the initial vertical position of θ = 0◦.

The pendulum drags with it a pointer that stops at the limit of the swing and identifies the value of θon a degree scale concentric with the pivot. The pendulum then free falls back to the vertical position tostop against the barrel.

A small amount of friction between the pointer and the scale prevents the pointer from falling backalong with the pendulum. The effect of this friction or the mass of the pointer on the system is negligible.

? The pointer, initially at rest, is accelerated along with the the pendulum arm on impact. Could itkeep moving past the limit of the pendulum arm, after the arm has stopped, and thus give inaccurateangle readings?

Data gathering and analysis

To determine the physical characteristics of the ballistic pendulum apparatus:

1. remove the pendulum arm from the ballistic pendulum assembly by unscrewing the pivot screw.Replace the screw for safekeeping;

29

2. verify that the plastic nut securing the brass weights to the bottom of the pendulum arm is tight;

3. measure with a digital scale (σ = ±0.01g) the mass of the ball m and ball/pendulum assembly MT ;

m = .............± ............. kg MT = .............± ............. kg

4. determine the centre of mass point of the pendulum/ball combination as shown in Figure 4.2. Makesure that the arm of the pendulum is parallel with the length of the scale.

? The pendulum placed on the edge of the measuring apparatus will be unstable and not remainbalanced in that position. How then can you precisely determine the length Rcm?

5. The centre-of-mass distance Rcm, the distance from the edge to the centre of the pivot hole on thearm of the pendulum of mass MT , is

Rcm = .............± ............. m.

Figure 4.2: Experimental setup for determining the centre-of-mass of the pendulum

6. With the launcher discharged and the ball removed, use the scale on the plunger to determine theoffset depth of the face of the piston from the front end of the barrel. You will need to subtract thisoffset from all of the following depth measurements. The offset is

offset = .............± ............. m.

7. With the ball removed, compress the piston until it latches at the short range setting. Measure thedepth from the face of the piston to the front end of the barrel. Subtract from this length the offsetdepth of the piston and record the result below and as x in Table 4.2:

xshort = .............± ............. m.

8. Repeat the above step for the medium and long range settings:

xmedium = .............± .............m xlong = .............± .............m

9. then determine the measurement error in the angle θ of the protractor scale used by the launcher

∆θ = ±.............◦


10. and finally, replace the pendulum arm, making sure that the pivot screw is tight.

Before you begin to gather ballistic data remember to avoid sitting directly in front of the discharge pathof the launcher barrel and

CAUTION: Always wear safety glasses while using the launcher.

1. Verify that the C-clamp securing the pendulum apparatus to the table is tight.

? What do you suppose might happen if the pendulum apparatus was discharged while not beingsecured to the table? How would your measurement be affected? What physical law is at work here?

2. Swing the pendulum to 90◦ and support it in that position with one hand or someone’s assistance.

3. With the other hand, load the ballistic pendulum by placing the projectile in the barrel and pushingit with the plunger rod until the trigger locks. This is the short range setting. Further compressionselects the medium range and finally, the long range setting.

4. Once the launcher is loaded, slowly withdraw the plunger, making sure that the trigger is indeedlocked and that the projectile has not rolled away from the face of the piston, as this would cause aninnacurate firing of the launcher.

5. Gently lower the pendulum to the vertical position, and move the angle indicator to the 0◦ mark. Ifthe indicator does not reach zero, you will need to subtract the offset from all your angle readings.

6. To fire the launcher, gently pull the string upward to release the trigger.

7. Perform five short range launches, recording the angle θi reached in trial i = 1 . . . 5 in the appropriatespaces of Table 4.1. These values should be within ± 0.5◦ of one another, otherwise redo the set ofmeasurements.

8. Perform five trials using the medium and then the long range settings.

☛✡

✟✠! Have a TA check and approve your angle data before proceeding with the calculations.

range θ1 θ2 θ3 θ4 θ5 〈θ〉 ∆θ

short

medium

long

Table 4.1: Experimental angle values at three force settings

• Calculate an average value 〈θ〉 for the five angles θi obtained in each of the three sets of trials andenter these in Tables 4.1 and 4.2. To avoid some lengthy standard deviation calculations, let theerror in the angle θi be the measurement error of the angle scale, ∆θ = ± 0.25◦.

? Looking at your data, is this shortcut a valid way of estimating the error in θ?

31

You now need to calculate v and ∆v for the three range settings. Note that in Equation 4.5 only the√

(1 − cos θ) term changes with θ. Chances of error will be minimized if the constant quantities areequated to a term C and ∆C and are evaluated only once. All the angle error values must be expressed inradians.

C =MT

m

√

2gRcm∆C

C=

√

(

∆MT

MT

)2

+

(

∆m

m

)2

+

(

∆Rcm

2Rcm

)2

............................... ......................................................

............................... ......................................................

vs = C√

(1− cos θ) ∆vs = |vs|

√

(

∆C

C

)2

+

(− sin θ ∆θ

2 cos θ

)2

............................... ......................................................

............................... ......................................................

vs = ....................± .................. m/s

? What should be the dimensions of C? And those of ∆C/C?

• Calculate the maximum kinetic energy of the steel ball at the moment that it lost contact with thepiston of the launcher and enter the value in Table 4.2. Show a complete calculation for the short

range setting:

Ks =1

2mv2s ∆Ks = |Ks|

√

(

∆m

m

)2

+

(

2∆vsvs

)2

............................... ......................................................

............................... ......................................................

Ks = ....................± .................. J

If no energy is lost (or gained) during the interaction, the kinetic energy K is equal to the potentialenergy V = kx2/2 of the spring before the ball was discharged, K = kx2/2. This is the equation of astraight line Y = MX with Y = K, X = x2 and slope M = k/2. Plotting K as a function of x2 yieldsfrom the slope a value for the spring constant k of the launcher spring.

• Shift focus to the Physicalab software and enter in the data window the three data pairs and corre-sponding errors as four space-delimited numbers: x2 K ∆K ∆x2.


range 〈θ〉 ◦ v (m/s) x (m) x2 (m2) K (J)

short ± ± ± ± ±medium ± ± ± ± ±long ± ± ± ± ±

Table 4.2: Parameters for the calculation of the kinetic energy K and the force constant k

• Select scatter plot. Click Draw to generate a graph of your data. Your graphed points shouldwell approximate a straight line.

☛✡

✟✠! Unless the three points are reasonably collinear, the fit will not yield a valid result. You need todetermine and correct the source of the error before proceeding. Consult the TA if you are stuck.

• Select fit to: y= and enter A*x+B in the fitting equation box. Click Draw to perform a linear

fit of the data. Label the axes and include a descriptive title. Click Send to: to email yourself acopy of the graph for later inclusion in your lab report.

• Summarize the values for the slope, the Y-intercept Y(X=0) and their associated errors, then calculatea value for k and ∆k:

slope = ............± ............ Y(X=0) = ............± ............

k = ..................... = ..................... = .....................

∆k = ..................... = ..................... = .....................

k = ............± ............

? The spring constant k is typically expressed in units of Newtons per metre. Using dimensionalanalysis, verify that your dimensions for k obtained from the graph agree with those of N/m.

• From the slope and Y-intercept, calculate the corresponding X-intercept at Y=0:

X = ..................... = ..................... = .....................

∆X = ..................... = ..................... = .....................

X = ............± ............m2.

33

• From this result estimate the spring preload distance x0:

x0 = ..................... = ..................... = .....................

∆x0 = ..................... = ..................... = .....................

x0 = ............± ............m.


Lab report


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(ta initials)


Experiment 5

Harmonic motion


• that Hooke’s Law F = −kx can be used to model the interaction of countless physical systems

• the basics of the Harmonic Oscillator, a cornerstone model in the analysis of physical problems

• that harmonic motion is esentially the movement of an object around the circumference of a circle

• this model cam easily be described by the rules of trigonometry

• how to set and check the calibration of a data gathering device;



Prelab preparation


To perform this Experiment and the Webwork Prelab Test successfully you need to be familiar with thecontent of this document and that of following FLAP modules (www.physics.brocku.ca/PPLATO). Beginby trying the fast-track quiz to gauge your understanding of the topic and if necessary review the modulein depth, then try the exit test. Check off the box when a module is completed.


FLAP MATH 1-5: Exponential and logarithmic functions

FLAP MATH 1-6: Trigonometric functions

Webwork: the Prelab Harmonic motion Test must be completed before the lab session☛✡


☛✡



34

35

The simple harmonic oscillator

A simple harmonic oscillator (SHO) is a model system that is widely used, and not just in the elementaryphysics exercises.The reason for this is profound: the same fundamental equations describe the motion ofa mass-on-a-spring, and also of the interatomic forces that hold all matter together. Literally, everythingwe touch, at the atomic level, is held together by springs connecting pairs of atoms. The details of theseinteractions are studied in Quantum Mechanics and Solid-State Physics, but two masses connected by acoiled elastic wire represent an excellent model system, from which much can be learned.

With no external forces applied to the material, the interatomic springs are at their equilibrium length,neither stretched nor compressed. The application of an external stretching force to the material will causethese springs to extend, thereby increasing the bulk length of the material. When the applied external forceis removed, the springs return to their equilibrium lengths, restoring the material to its original dimensions.

This restoring force may be overcome by a large enough applied external force that will cause the objectto deform permanently or to break. The maximum force applicable without permanent distortion is calledthe elastic limit of the material.

Hooke’s Law states that the amount of stretch or compression x exhibited by a material is directlyproportional to the applied force F . The proportionality constant, called the spring constant k, has theunits of Newtons per metre (N/m) and is a measure of the stiffness of the material being considered. Thisproportionality between force and elongation has been found to hold true for any body as long as theelastic limit of the material is not exceeded.

Perhaps the most convenient way of expressing a relationship of the Hooke’s Law is to write;

F = −kx, (5.1)

The minus sign expresses the fact that the force exerted by the spring, Fs, is always opposing the defor-mation. It is often described as a restoring force

If Fs is the only force acting on the system, the system is called a simple harmonic oscillator, and itundergoes simple harmonic motion: sinusoidal oscillations about the equilibrium point, with a constantamplitude and a constant frequency, which does not depend on the amplitude. Harmonic motion isequivalent to an object moving around the circumference of a circle at constant speed.

In the absence of friction, the oscillations will conrinue forever. Friction robs the oscillator of it’s energy,the oscillations decay, and eventually stop altogether. Frequently, this friction term is proportional to thevelocity of the moving mass (think of the air resistance), Fd = Rv.

If a frictional force Fd = −Rv, proportional and opposed to the velocity is also present, the harmonicoscillator is described as a damped oscillator. Depending on the strength of the friction coefficient R, thesystem can either oscillate with a frequency slightly smaller than in the non-damped case, and an amplitudedecreasing with time (underdamped oscillator); or decay exponentially to the equilibrium position, withoutoscillations (overdamped oscillator).

Mass on a vertical spring

When a mass m is attached to a hanging spring, the spring is subjected to a force Fg = mg and will bestretched to a new equilibrium position y0 where

Fg = −Fs ⇒ mg = ky0 ⇒ m =k

gy0. (5.2)

If the mass is displaced from y0 and released, it will begin to oscillate about y0 according to

y = A0 cos(ω0t+ φ), ω0 =√

k/m = 2πf0 =2π

T0(5.3)

36 EXPERIMENT 5. HARMONIC MOTION

where A0 and φ are the initial amplitude and phase angle of the oscillation, T0 is the period in seconds,and ω0 is the angular speed in radians/second. If a damping force Fd is present, the oscillation decaysexponentially at a rate determined by the damping coefficient γ:

y = A0 cos(ωdt+ φ)e(−γt), γ =R

2m, ωd =

√

ω20 − γ2 (5.4)

Note two interesting details; first, the damped frequency of oscillations, ωd, is made smaller than ω0 bythe subtraction of γ2 under the square root. However, if R is not very large, this reduction is not allthat noticeable, even though the decrease in the amplitude due to the e−γt term may be readily observed.Second, in the case of the air resistance, Fd = −Rv is an approximate relation, valid only for slow speeds.As the speed increases, the turbulent air flow offers resistance that is proportional to the square of thevelocity, Fd = −Cv2. Here C is the so-called drag coefficient. Unfortunately, this expression cannot besolved analytically and would have to use numerical methods (computer) to evaluate. Note also thatthe hanging spring is stretched by its own weight and may exhibit twisting as well as lateral oscillationswhen stretching. One other simplifying assumption: this experiment assumes an ideal massless springconnected to a point mass m. Even with all these approximations, Equation 5.4 lends itself very nicely toan experimental investigation.

Procedure and analysis

The experimental setup consists of a vertical stand from which hangs a spring. A platform attached to thebottom of the spring accepts various mass loads. The spring-mass system is free to move in the verticaldirection. When the platform is static, it’s height above the table can be measured with a ruler. Whenthe system is in motion, Physicalab can be used to record the time-dependent change in distance from arangefinder to the bottom of the platform. The rangefinder measures distance by emitting an ultrasonicpulse and timing the delay for the echo to return.

Part 1: Static properties of springs

In the following exercises you will determine the force constants of two different springs, distinguished as#1 and #2. You can then verify that their combined force constant satisfies the equation derived in thereview section of the experiment. Note: Use positive k values in all the following steps.

Equation 5.2 represents a straight line with slope k/g = ∆m/∆y so that k = |∆m/∆y| ∗ g. By varyingm, measuring y0 and fitting the resulting data to the equation of a straight line, the force constant k forthe spring can be determined. Note that the absolute distance y0 does not matter, however the change indistance ∆y with mass ∆m is critical.

• Suspend spring #1 from the holder and load it with the 50 g platform. Use a ruler to carefullymeasure the distance y from the top of the table to the bottom of the platform to a precision of1 mm. Record your result in Table 5.1.

• Determine the mass required to bring the platform near the top of the table and record this newdistance, then select several intermediate masses and fill in Table 5.1.

? Why do you want to determine a mass that brings the platform near to the table?

• Enter the data pairs (y,m) in the Physicalab data window. Select scatter plot and click Draw togenerate a graph of your data.

37

mass (kg)

y (m)

Table 5.1: Data for determining the spring force constant k1 of spring #1

? Did you choose to include the mass of the platform (50g) as part of the stretching massm in Table 5.1?How would the graph and the resulting value for the spring constant k be affected by your choice?

• Select fit to: y= and enter A+B*x in the fitting equation box. Click Draw to perform a linear

fit on your data, then evaluate k1 and the associated error σ(k1). Click Send to: to email yourselfa copy of the graph for later inclusion in your lab report.

B = ........... ± ........... A = ........... ± ...........

k1 = ..................... = ..................... = .....................

∆k1 = ..................... = ..................... = .....................

k1 = ............... ± ...............

• Repeat the above procedure using spring #2, recording your results in Table 5.2, then determine thespring constant k2 for the spring.

mass (kg)

y (m)

Table 5.2: Data for determining the spring force constant k2 of spring #2

B = ........... ± ........... A = ........... ± ...........

k2 = ..................... = ..................... = .....................

∆k2 = ..................... = ..................... = .....................

k2 = ............... ± ...............

• Connect together springs #1 and #2, and repeat the procedure, entering your data in Table 5.3, todetermine the experimental effective spring constant ke of the two springs:

B = ........... ± ........... A = ........... ± ...........


mass (kg)

y (m)

Table 5.3: Data for determining the effective spring force constant ke of springs #1 and #2

ke = ..................... = ..................... = .....................

∆ke = ..................... = ..................... = .....................

ke = ............... ± ...............

• Calculate the theoretical effective spring constant ke and ∆ke for the two springs #1 and #2:

k1 = ........... ± ........... k2 = ........... ± ...........

ke =k1k2

k1 + k2= ..................... = .....................

∆ke = k2e

√

(

∆k1k21

)2

+

(

∆k2k22

)2

= ..................... = .....................

ke = ............... ± ...............

Part 2: Damped harmonic oscillator

You will now explore the behaviour in time of an oscillating mass with the aid of a computer-controlledrangefinder. This device sends out an ultrasonic pulse that reflects from an object in the path of the conicalbeam and returns to the rangefinder as an echo. The rangefinder measures the elapsed time to determinethe distance to the object, using the speed of sound at sea level as reference.

By accumulating a series of coordinate points (ω,m) and fitting your data to Equation 5.4, you canuse several methods to determine the spring constant for the oscillating mass system. Begin by selectingone of the two springs to use:

• With each of the masses used previously with this spring, raise the platform a small distance, thenrelease it to start the mass swinging vertically. Wait until the spring/mass system no longer exhibitsany erratic oscillations.

• Login to the Physicalab software. Check the Dig1 box and choose to collect 200 points at 0.05

s/point. Click Get data to acquire a data set.

• Select scatter plot, then Click Draw . Your points should display a smooth slowly decaying sinewave, without peaks, stray points, or flat spots. If any of these are noted, adjust the position of therangefinder and acquire a new data set.

• Select fit to: y= and enter A*cos(B*x+C)*exp(-D*x)+E in the fitting equation box. Click

Draw . If you get an error message the initial guesses for the fitting parameters may be too distant

39

from the required values for the fitting program to properly converge. Look at your graph and entersome reasonable initial values for the amplitude A of the wave and the average (equilibrium) distanceE of the wave from the detector. C corresponds to the initial phase angle of the sine wave at x=0.

• The angular speed B (in radians/s) is given by B= 2π/T . Estimate the time x= T between twoadjacent minima of the sine wave then estimate and enter an initial guess for B.

• The damping coefficient D determines the exponential decay rate of the wave amplitude to theequilibrium distance E. When D = 1/x, the envelope will have decreased from A0 to A0e

−1 =A0/e = A0/2.718 ≈ A0/3. Make an initial guess for D by estimating the time t=x required for theenvelope to decrease by 2/3.

• Check that the fitted waveform overlaps well your data points, then label the axes and include aspart of the title the value of mass m used. Email yourself a copy of all graphs as part of your labdata set.

• Record the results of the fit in Table 5.4, then complete the table as before.

mass (kg)

y0 (m)

ω0 (rad/s)

γ (s−1)

Table 5.4: Experimental results for damped harmonic oscillator

• The spring constant k can be determined as before from the slope of a line fitted through the (y0,m)data in Table 5.4. Generate and save the graph, then record the result below:

k(y0) = ............... ± ...............

• The spring constant k can also be determined fron the period of oscillation of the mass. Rearrangingthe terms in Equation 5.3 yields m = k/ω2

0 . Enter the coordinate pairs from Table 5.4 in the form(ω0,m) and view the scatter plot of your data. Select fit to: y= and enter A+B/x**2 in thefitting equation box. This is more convenient than squaring all the ω0 values and fitting to A+B/x.

Click Draw to perform a quadratic fit on your data. Print the graph, then enter below the valuefor the spring constant:

k(ω0) = ............... ± ...............

• The damped harmonic oscillator equation 5.4 predicts that the frequency of oscillation depends on thedamping coefficient γ so that your experiment actually yields values for ωd rather than ω0. Calculateω0 from ωd using an average value for γ.

? Which ωd value should be used? Why?

ω0 = ..................... = ..................... = .....................

? Is the difference between ωd and ω0 significant? Explain.


Calibration of rangefinder

It was assumed that the rangefinder is properly calibrated and measuring the correct distance. It is alwaysa good practice to check the calibration of instrument scales against a known good reference, in thiscase the ruler scale. Calibration errors are systematic errors and can be corrected. Once the amount ofmis-calibration is determined, the mis-calibrated data can be adjusted to represent correct values.

• Check the calibration of your rangefinder against the ruler. Compare the distance results for twopoints from the (y0,m) data in Part 1 (ruler) and Part 2 (rangefinder). The mass difference betweenthe points must be the same for both parts. If the rangefinder is calibrated, the distance differencesshould also be the same (within measurement error).

? How do you choose the two calibration points that give the least relative error?

Part 1: y1 = ......... y2 = ......... y2 − y1 = ......... ± .........

Part 2: y1 = ......... y2 = ......... y2 − y1 = ......... ± .........

? Why are two calibration points sufficient?


Lab report


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Appendix A

Review of math basics

Fractions

a

c+

b

d=

ad+ bc

cd; If

a

c=

b

d, then ad = cb and

ad

bc= 1.

Quadratic equations

Squaring a binomial: (a+ b)2 = a2 + 2ab+ b2

Difference of squares: a2 − b2 = (a+ b)(a− b)

The two roots of a quadratic equation ax2 + bx+ c = 0 are given by x =−b±

√b2 − 4ac

2a.

Exponentiation

(ax)(ay) = a(x+y) ,ax

ay= ax−y , a1/x = x

√a , a−x =

1

ax, (ax)y = a(xy)

Logarithms

Given that ax = N , then the logarithm to the base a of a number N is given by logaN = x.For the decimal number system where the base of 10 applies, log10 N ≡ logN and

log 1 = 0 (100 = 1)

log 10 = 1 (101 = 10)

log 1000 = 3 (103 = 1000)

Addition and subtraction of logarithms

Given a and b where a, b > 0: The log of the product of two numbers is equal to the sum of the individuallogarithms, and the log of the quotient of two numbers is equal to the difference between the individuallogarithms .

log(ab) = log a+ log b

log(a

b

)

= log a− log b

The following relation holds true for all logarithms:

log an = n log a

42

43

Natural logarithms

It is not necessary to use a whole number for the logarithmic base. A system based on “e” is often used.Logarithms using this base loge are written as “ln”, pronounced “lawn”, and are referred to as natural

logarithms. This particular base is used because many natural processes are readily expressed as functionsof natural logarithms, i.e. as powers of e. The number e is the sum of the infinite series (with 0! ≡ 1):

e =

∞∑

n=0

1

n!=

1

0!+

1

1!+

1

2!+

1

3!+ · · · = 2.71828 . . .

Trigonometry

Pythagoras’ Theorem states that for a right-angled triangle c2 = a2 + b2.Defining a trigonometric identity as the ratio of two sides of the triangle,there will be six possible combinations:

sin θ =b

ccos θ =

a

ctan θ =

b

a=

sin θ

cos θ

csc θ =c

bsec θ =

c

acot θ =

a

b=

cos θ

sin θ

sin(θ ± φ) = sin θ cosφ± cos θ sinφ sin 2θ = 2 sin θ cos θ 180◦ = π radians = 3.15159 . . .

cos(θ ± φ) = cos θ cosφ∓ sin θ sinφ cos 2θ = 1− 2 sin2 θ 1 radian = 57.296 . . .◦

tan(θ ± φ) =tan θ ± tanφ

1∓ tan θ tan φtan 2θ =

2 tan θ

1− tan2 θsin2 θ + cos2 θ = 1

To determine what angle a ratio of sides represents, calculate the inverse of the trig identity:

if sin θ =b

c, then θ = arcsin

(

b

c

)

For any triangle with angles A,B,C respectively opposite the sides a, b, c:

a

sinA=

b

sinB=

c

sinC, (sine law) c2 = a2 + b2 − 2ac cosC. (cosine law)

The sinusoidal waveform

Consider the radius vector that describes the circumference of a circle, as shown in Figure A.1 If we increaseθ at a constant rate from 0 to 2π radians and plot the magnitude of the line segment b = c sin θ as afunction of θ, a sine wave of amplitude c and period of 2π radians is generated.

Relative to some arbitrary coordinate system, in this case the X-Y axis shown, the origin of this sinewave is located at a offset distance y0 from the horizontal axis and at a phase angle of θ0 from the verticalaxis.

The sine wave referenced from this (θ, y) coordinate system is given by the equation

y = y0 + c sin(θ + θ0)

44 APPENDIX A. REVIEW OF MATH BASICS

Figure A.1: Projection of a circular motion to an X-Y plane to generate a sine wave

Appendix B

Error propagation rules

• The Absolute Error of a quantity Z is given by ∆Z, always ≥ 0.

• The Relative Error of a quantity Z is given by ∆ZZ , always ≥ 0.

• If a constant k has no error associated with it: constant factors out of relative error

Z = kA ∆Z = k∆A and∆Z

Z=

∆A

A

• Addition and subtraction: note that error terms always add

Z = kA±B ± . . . ∆(Z) =√

(k∆A)2 + (∆B)2 + . . .

• Multiplication and division: constants factor out of relative errors

Z =kA×B × . . .

C ×D × . . .

∆Z

Z=

√

(

∆A

A

)2

+

(

∆B

B

)2

+

(

∆C

C

)2

+

(

∆D

D

)2

. . .

• Functions of one variable: if the quantity A is measured with uncertainty ∆A and is then used tocompute F (A), then the uncertainty ∆F in the value of F (A) is given by

∆F =

(

dF

dA

)

∆A

Function F (A) Derivative, dFdA Error equation

An nAn−1 ∆FF = n∆A

A

logeA A−1 ∆F = ∆AA

exp(A) exp(A) ∆FF = ∆A

sin(A) cos(A) ∆F = cos(A)∆A

cos(A) − sin(A) ∆F = − sin(A)∆A

tan(A) sec(A)2 ∆F = sec(A)2∆A

All trigonometric functions and the errors in the angle variables are evaluated in radians

45

46 APPENDIX B. ERROR PROPAGATION RULES

How to derive an error equation

Let’s use the change of variable method to determine the error equation for the following expression:

y =M

m

√

0.5 kx (1− sin θ) (B.1)

• Begin by rewriting Equation B.1 as a product of terms:

y = M ∗ m−1 ∗ [ 0.5 ∗ k ∗ x ∗ (1− sin θ)] 1/2 (B.2)

= M ∗ m−1 ∗ 0.51/2 ∗ k1/2 ∗ x1/2 ∗ (1− sin θ)1/2 (B.3)

• Assign to each term in Equation B.3 a new variable name A,B,C, . . . , then express v in terms ofthese new variables,

y = A ∗ B ∗ C ∗ D ∗ E ∗ F (B.4)

• With ∆(y) representing the error or uncertainty in the magnitude of y, the error expression for y iseasily obtained by applying Rule 4 to the product of terms Equation B.4:

∆(y)

y=

√

(

∆(A)

A

)2

+

(

∆(B)

B

)2

+

(

∆(C)

C

)2

+

(

∆(D)

D

)2

+

(

∆(E)

E

)2

+

(

∆(F )

F

)2

(B.5)

• Select from the table of error rules an appropriate error expression for each of these new variables asshown below. Note that F requires further simplification since there are two terms under the squareroot, so we equate these to a variable G:

A = M , ∆(A) = ∆(M)

B = m−1,∆(B)B = −1

∆(m)m = −∆(m)

m

C = 0.51/2,∆(C)C = 1

2∆(0.5)|0.5| = 0

D = k1/2,∆(D)D = 1

2∆(k)k

=∆(k)2k

E = x1/2,∆(E)E = 1

2∆(x)x =

∆(x)2x

F = G1/2,∆(F )F = 1

2∆(G)G =

∆(G)2G

G = 1− sin θ, ∆(G) =√

(∆(1))2 + (∆(sin θ))2 = cos θ∆θ

• Finally, replace the error terms into the original error Equation B.5, simplify and solve for ∆(y) bymultiplying both sides of the equation with y:

∆(y) = y

√

(

∆M

M

)2

+

(

∆m

m

)2

+

(

∆k

2k

)2

+

(

∆x

2x

)2

+

(

cos θ∆θ

2− 2 sin θ

)2

(B.6)

Appendix C

Graphing techniques

Figure C.1: Proper scaling of axes

Figure C.2: Improper scaling of axes

A mathematical function y = f(x) describes the one to onerelationship between the value of an independent variable xand a dependent variable y. During an experiment, we analysesome relationship between two quantities by performing a se-ries of measurements. To perform a measurement, we set somequantity x to a chosen value and measure the correspondingvalue of the quantity y. A measurement is thus representedby a coordinate pair of values (x, y) that defines a point on atwo dimensional grid.

The technique of graphing provides a very effective methodof visually displaying the relationship between two variables.By convention, the independent variable x is plotted along thehorizontal axis (x-axis) and the dependent variable y is plottedalong the vertical axis (y-axis) of the graph. The graph axesshould be scaled so that the coordinate points (x, y) are welldistributed across the graph, taking advantage of the maxi-mum display area available. This point is especially impor-tant when results are to be extracted directly form the datapresented in the graph. The graph axes do not have to startat zero.

Scale each axis with numbers that represent the range ofvalues being plotted. Label each axis with the name and unitof the variable being plotted. Include a title above the graph-ing area that clearly describes the contents of the graph beingplotted. Refer to Figure C.1 and Figure C.2.

The line of best fit

Suppose there is a linear relationship between x and y, so thaty = f(x) is the equation of a straight line y = mx + b where m is the slope of the line and b is the valueof y at x = 0. Having plotted the set of coordinate points (x, y) on the graph, we can now extract a valuefor m and b from the data presented in the graph.

Draw a line of ’best fit’ through the data points. This line should approximate as well as possible thetrend in your data. If there is a data point that does not fit in with the trend in the rest of the data, youshould ignore it.

47

48 APPENDIX C. GRAPHING TECHNIQUES

The slope of a straight line

Figure C.3: Slope of a line

The slope m of a straight-line graph is determined bychoosing two points, P1 = (x1, y1) and P2 = (x2, y2),on the line of best fit, not from the original data, andevaluating Equation C.1. Note that these two pointsshould be as far apart as possible.

m =rise

run

m =∆y

∆x=

y2 − y1x2 − x1

(C.1)

Error bars

Figure C.4: Error Bars for Point (x, y)

All experimental values are uncertain to some degree dueto the limited precision in the scales of the instrumentsused to set the value of x and to measure the result-ing value of y. This uncertainty σ of a measurement isgenerally determined from the physical characteristics ofthe measuring instrument, i.e. the graduations of a scale.When plotting a point (x, y) on a graph, these uncertain-ties σ(x) and σ(y) in the values of x and y are indicatedusing error bars.

For any experimental point (x± σ(x), y ± σ(y)), theerror bars will consist of a pair of line segments of length2σ(x) and 2σ(y), parallel to the x and y axes respec-tively and centered on the point (x, y). The true value lies within the rectangle formed by using the errorbars as sides. The rectangle is indicated by the dotted lines in Figure C.3. Note that only the error bars,and not the rectangle are drawn on the graph.

The uncertainty in the slope

Figure C.5: Determining slope error

Figure C.5 shows a set of data points for a linear rela-tionship. The slope is that of line 2, the line of best fitthrough these points. The uncertainty in this slope istaken to be one half the difference between the line ofmaximum slope line 1 and the line of miinimum slope,line 3:

σ(slope) =slopemax − slopemin

2(C.2)

The lines of maximum and minimum slope should gothrough the diagonally opposed vertices of the rectanglesdefined by the error bars of the two endpoints of thegraph, as in Figure C.5.

49

Logarithmic graphs

In science courses you will encounter a great number of functions and relationships, both linear and non-linear. Linear functions are distinguished by a proportional change in the value of the function with a changein value of one of the variables, and can be analyzed by plotting a graph of y versus x to obtain the slope mand vertical intercept b. Non-linear functions do not exhibit this behaviour, but can be analyzed in a similarmanner with some modification. For example, a commonly occuring function is the exponential function,

Figure C.6: The exponential function y = aebx.

y = aebx, (C.3)

where e = 2.71828 . . ., and a and b are constants.Plotted on linear (i.e. regular) graph paper, thefunction y = aebx appears as in Figure C.6. Tak-ing the natural logarithm of both sides of equa-tion (C.3) gives

ln y = ln(

aebx)

ln y = ln a+ ln(

ebx)

ln y = ln a+ bx ln e

ln y = ln a+ bx (since ln e = 1)

Equation (C.4) is the equation of a straight linefor a graph of ln y versus x, with ln a the verticalintercept, and b the slope. Plotting a graph of ln yversus x (semilogarithmic, i.e. logarithmic on the vertical axis only) should result in a straight line, whichcan be analyzed.

There are two ways to plot semilogarithmic data for analysis:

1. Calculate the natural logarithms of all the y values, and plot ln y versus x on linear scales. The slopeand vertical intercept can then be determined after plotting the line of best fit.

2. Use semilogarithmic graph paper. On this type of paper, the divisions on the horizontal axis areproportional to the number plotted (linear), and the divisions on the vertical axis are proportional tothe logarithm of the number plotted (logarithmic). This method is preferable since only the naturallogarithms of the vertical coordinates used to determine the slope of the lines best fit, minimum andmaximum slope need to be calculated.

Semilogarithmic graph paper

The horizontal axis is linear and the vertical axis is logarithmic. The vertical axis is divided into a seriesof bands called decades or cycles.

• Each decade spans one order of magnitude, and is labelled with numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 1.

• the second “1” represents 10× what the first “1” does, the third 10× the second, et cetera, and

• there is no zero on the logarithmic axis since the logarithm of zero does not exist.

A logarithmic axis often has more than one decade, each representing higher powers of 10. In Figure C.7,the axis has 3 decades representing three consecutive orders of magnitude. For instance, if the data to beplotted covered the range 1 → 1000, the lowest decade would represent 1 → 10 (divisions 1, 2, 3, . . . , 9),

50 APPENDIX C. GRAPHING TECHNIQUES

Figure C.7: A 3-Decade Logarithmic Scale.

the second decade 10 → 100 (divisions 10, 20, 30, . . . , 90) and the third decade 100 → 1000 (divisions 100,200, 300, . . . 900, 1000).

Another advantage of using a logarithmic scale is that it allows large ranges of data to be plotted. Forinstance, plotting 1 → 1000 on a linear scale would result in the data in the lower range (e.g. 1 → 100)being compressed into a very small space, possibly to the point of being unreadable. On a logarithmicscale this does not occur.

Calculating the slope on semilogarithmic paper

The slope of a semilogarithmic graph is calculated in the usual manner:

m = slope

=rise

run

=∆(vertical)

∆(horizontal).

For ∆(vertical) it is necessary to calculate the change in the logarithm of the coordinates, not the changein the coordinates themselves. Using points (x1, y1) and (x2, y2) from a line on a semilogarithmic graph ofy versus x and Equation C.4, the slope of the line is obtained.

m =rise

run

m =ln y2 − ln y1x2 − x1

m =ln (y2/y1)

x2 − x1(C.4)

Note that the units for m will be (units of x)−1 since ln y results in a pure number.

Analytical determination of slope

There are analytical methods of determining the slope m and intercept b of a straight line. The advantageof using an analytical method is that the analysis of the same data by anyone using the same analyticalmethod will always yield the same results. Linear Regression determines the equation of a line of best fitby minimizing the total distance between the data points and the line of best fit.

To perform “Linear Regression” (LR), one can use the preprogrammed function of a scientific calculatoror program a simple routine using a spreadsheet program. Based on the x and y coordinates given to it, aLR routine will return the slope m and vertical intercept b of the line of best fit as well as the uncertaintiesσ(m) and σ(b) in these values. Be aware that performing a LR analysis on non-linear data will producemeaningless results. You should first plot the data points and determine visually if a LR analysis is indeedvalid.

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