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DATA ANALYSIS WORKSHEET
Introduction This worksheet follows on from the Errors and Accuracy lecture and contains a series of problems/exercises which involve (a) calculating formulae for errors, (b) calculating values for errors, (c) using the line fitting programme. These exercises include the use of the Excel spreadsheet on the PC's to process some of these error values. The exercises also involve (d) using the spreadsheet to solve "problems" which might occur in other areas of your course (for example to numerically solve an equation).
a) Calculating Formulae for Error Bars The basic formulae that you need here are all given in the two tables in the Errors and Accuracy chapter of the Lab. Manual, you just need to apply them to the cases given.
(i) The cross sectional area of a (circular) piece of wire is given by A=πr 2. If we have measured the diameter of the piece of wire as d±∆d what is the formula for the error in the area ∆A in terms of the error in the diameter ∆d?
Ans:
If the value of d = 1.22±0.02 mm, what is the value for A =
(ii) The frequency of oscillation f of a pendulum is given in terms of the periodic time of the pendulum T by the formula T = 1 / f. If the measured periodic time has an error value of ∆T calculate the formula for the error bar in the frequency ∆f in terms of ∆T.
Ans:
If the periodic time of the pendulum is 2.3±0.1s what is the value and error bar of the frequency f=
(iii) Two resistors R 1 and R 2 are added together in series to give a total resistance RT = R 1 + R 2. If resistor R 1 has an error bar ∆R 1 and resistor R 2 has an error bar ∆R 2 what is the formula for the error bar of the total resistance RT?
Ans:
If the measured value and error bar for R 1 = 10.3±0.2Ω and for resistor R 2 = 5.5±0.4Ω what is the value and error bar for R (^) T =
(iv) The decay time T of a capacitor of capacitance C through a resistor of resistance R is given by the formula T = R C. If the error in R is ∆R and the error in C is ∆C then what is the formula for the error ∆T in the decay time?
Ans:
If the value of the resistor is R = 1000± 50 Ω and the value of the capacitance is C = 10 ±1nF what is the value of the decay time T=
(v) The decay time T of an inductor of inductance L through a resistor of resistance R is given by T = L / R. If the error in the value of the inductance is ∆L and the error in the value of the resistance is ∆R what is the formula for the error in the decay time ∆T?
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If the value of the inductance is L = 100±5mH and the value of the resistance is R = 50 ± 1 Ω then what is the value of the decay time T =
(vi) For a diffraction grating the wavelength of light is given by the formula λ = d sinθ where d is the distance between rulings on the grating and θ is the angle at which the diffraction line is measured. If ∆d is the error in the distance between rulings and ∆θ is the error in the angle measurement what is the formula for the error in the wavelength ∆λ?
Ans:
If the values of d and θ are given by d = 1.67±0.03μm and θ = 17.42±0.07° what is the value for λ =
[Note of caution here, think carefully about the units for the angle !]
Use LineFit to plot the data and perform a straight line fit. The equation which relates the effective value of g' to the height h is
g g
h RE
where g is the acceleration due to gravity at the surface of the Earth and RE is the radius of the Earth. From the slope and intercept of your straight line fit calculate values (and their error bars) for g and R (^) E.
Ans:
(ii) In problem (ii) of the Spreadsheet section you were given a set of data for the decay of the voltage across a capacitor with time. Plot a graph of that data using LineFit and perform a straight line fit.
Comment on the resulting fit:
Now using the results you obtained from exercise (iii) in the Spreadsheet section plot a graph of the Logarithm of the Voltage against time.
Comment on the resulting fit:
Using a Spreadsheet to Plot and Solve Equations A spreadsheet can be used as a very useful tool for plotting and numerically solving some equations/problems. Let's do an example first. Say we wanted to find the zeroes of the equation y = x^2 - 3x - 3. Clearly you could do this via the well known quadratic equation formula but let's do it numerically as an exercise;
First we estimate the range in which the equation will go through zero. It's numerically clear that if x is less than -4 and greater than 4 the x^2 term will always dominate and so we need only search the range -4 < x < 4.
In a blank spreadsheet enter in cell A1 the value -4.0. In cell A2 enter the formula =A1+0.2 and press return. Use the mouse to click on cell A2 and then highlight down to cell A41. Click on Edit and Fill Down. We should now have a column of numbers from -4 to +4. Click into cell B1 and enter the formula = A1^2 - 3*A1 - 3 and press return. Select cell B1 with the mouse and highlight down to B41. Click on Edit and then Fill Down.
If we now examine the column B we have a list of values for the function and can look to see where it crosses from positive to negative and hence goes through zero. We can plot a graph of these values as follows. Use the mouse to select and highlight the cells in columns A and B which contain values. Go to the top of the Excel toolbar and find the "Chart Wizard" (as you slowly pass over the symbols a name for the symbol should appear). Click on the chart wizard. Select an X-Y Scatter Diagram and then click on next>. Select the top of the right hand column of sub- charts which are displayed and then click on next>. Continue to click on next> until it becomes greyed out and then click on Finish. Hey presto, a graph should appear on your spreadsheet.
Say we want a more accurate numerical value than we have obtained so far. Well the equation crosses somewhere between -1.0 and -0.6 so go to cell A1 and change the value from -4. to -1.0. Then go to cell A2 and change the formula from =A1+0. to =A1+0.01, select A2 and fill down to A41. Observe what happens to the values in column B and the graph as you do this. We can, of course, do the same for the other root of the equation to obtain it more accurately.
(i) Find the two zeroes of the equation ex^ - 3x = 0 using the spreadsheet. The values lie within the range 0 < x < 2. Note although this is a much harder equation than the example the numerical technique with the spreadsheet is just the same. Find the solutions to an accuracy of ±0.01.
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(ii) Find the two zeroes of the equation cos(πx) = x^2 + x, they lie within the range -1. to 0.5. Find the solutions to an accuracy of ±0.02. (Note: π=3.141593)
Ans:
(iii) A slightly different problem. Plot a graph of the function
x x
3
exp( ) − 1
over the range
from x=0 to x=5 using the spreadsheet and then use the method of refining the x- values which we used in (i) and (ii) to find the maximum value of the function and the value of x at which it occurs.
Ans:
[Note: Be careful about the result for x=0, should you include or exclude it from the spreadsheet. Computers can't think, they'll just blindly do what you tell them even if that isn't sensible.]
