Measurements and errors

1. 1 ERROR MEASUREMENT A SHORT NOTES Arun Umrao https://sites.google.com/view/arunumrao DRAFT COPY - GPL LICENSING

2. 2 Errors Contents 1 Errors 3 1.1 Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.1 Estimates and Errors . . . . . . . . . . . . . . . . . . . 3 1.1.2 Accuracy & Precision . . . . . . . . . . . . . . . . . . . 4 1.1.3 Discrepancy . . . . . . . . . . . . . . . . . . . . . . . . 4 1.1.4 Acceptance & Measured Value . . . . . . . . . . . . . . 6 1.1.5 Why Errors are Absolute? . . . . . . . . . . . . . . . . 7 1.2 Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2.1 Calculation of Errors . . . . . . . . . . . . . . . . . . . 10 1.2.2 Limitation of Errors . . . . . . . . . . . . . . . . . . . 14 1.2.3 Error By Calculus . . . . . . . . . . . . . . . . . . . . . 15 1.2.4 Application of Error Analysis . . . . . . . . . . . . . . 23 1.2.5 Rounding Off Errors . . . . . . . . . . . . . . . . . . . 35 1.2.6 Error in Sample Measurements . . . . . . . . . . . . . 36 1.3 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3. 1.1. MEASUREMENTS 3 1Errors 1.1 Measurements Measurement is a process in which a physical quantities are measured with respect to some internationally accepted units. For example $1=Rs. 60/- . This is conversion of US dollar into Indian rupee. Here $ and Rs. are the currencies and 60 is conversion ratio. During measurement errors are generated and to find the accurate measurement, we have to remove errors and discrepancies. 1.1.1 Estimates and Errors 0 0 30 5 60 10 90 15 120 20 150 25 180 30 210 35 240 40 270 45 300 50 330 55 (a) 0 0 30 5 60 10 90 15 120 20 150 25 180 30 210 35 240 40 270 45 300 50 330 55 (b) Figure 1.1: Uncertainty in measurement of angle precise upto second value. Suppose a direction measuring instrument (compass) can measures the an- gular direction upto second unit. From above figure 1.1, red hand indicates the degree reading, blue hand minute reading and black hand second reading. Carefully observing the instrument the degree reading is zero, minute reading is 20 and second reading is between 15 to 16. 15 is the certain reading for second hand but fraction of second reading is uncertain. Hence the measured

4. 4 Errors reading of the instrument is 00 20′ 15.5′′ . Observer can certainly say that the certain reading is 00 20′ 15′′ but (s)he is uncertain about the fraction reading of second hand. This means first estimate reading of angle is 00 20′ 15′′ , as the angle is closer to it. There is no more precise reading possible, hence probable reading of the angle is 00 20′ 15′′ ± 0.5′′ . The best estimate reading is 00 20′ 15′′ . The probable reading is from 00 20′ 14.5′′ to 00 20′ 15.5′′ . Scientifically it can be written as 00 20′ 14.5′′ ≤ θ ≤ 00 20′ 15.5′′ If best precise reading of the observation is known then the deviation of observed reading with best precise reading is known as error. In second figure, if we remove the second hand of compass, then one can certainly say that the direction is 00 20′ . This value is more certain but have less precision. This leads that as the measuring unit increases, error and precision decreases. 1.1.2 Accuracy & Precision Accuracy of a measurement system is the degree of closeness of measurements of a quantity to the actual value. It depends on the measuring instruments. For example, Vernier Caliper is more accurate in measuring of wire thickness than student’s foot scale. Error of accuracy (say e) is given by e = true value − measured value The precision of a measurement system is the degree to which repeated measurements under unchanged conditions show the same results. It depends on the sample size. Precision (say ǫ) is given by ǫ = 1 σ2 1.1.3 Discrepancy Two students measure the value of resistance by using ohm’s law and they give different results. There is discrepancy in the results due to difference in result. In other words when two measurements of a same quantity are disagree with one another then there is a discrepancy. For example,

5. 1.1. MEASUREMENTS 5 two students measure the resistance value of the same conductor are (25±2)Ω and (20 ± 2)Ω respectively. The discrepancy in the results is D = 25 − 20 = 5Ω 15 16 17 0 1 2 O Ω b b 15 16 17 0 1 2 O Ω b b Figure 1.2: In first figure the observations do not fall within the domain of each other, hence the discrepancy is significant. While in second figure, observations fall within the domain of each other, hence discrepancy is not significant. Significance of Discrepancy The maximum and minimum possible resistance measured by first student are (25 + 2 = 27)Ω & (25 − 2 = 23)Ω. The maximum and minimum possible resistance measured by second student are (20 + 2 = 22)Ω & (20 − 2 = 18)Ω respectively. Maximum and minimum values of measured resistance of both student do not fall in one others domain, Hence discrepancy is significant. Again in second experiment, resistance value of same conductor are measured by both students are (25 ± 3)Ω and (20 ± 3)Ω respectively. The discrepancy in the results is D = 25 − 20 = 5Ω The maximum and minimum possible resistance measured by first student are 25 + 3 = 28Ω & 25 − 3 = 22Ω respectively. The maximum and minimum possible resistance measured by first student student are 20 + 3 = 23Ω & 20 − 3 = 17Ω respectively. Maximum and minimum values of measured resistances fall in one others domain, Hence discrepancy is not significant.

6. 6 Errors 1.1.4 Acceptance & Measured Value Accepted value of a physical quantity is that value which is widely accepted by all the persons and labs or taking global physical conditions. For example velocity of sound in air at normal temperature and pressure is 331m/s. Sim- ilarly, gas constant is accepted globally as (8.31451 ± 0.00007)J/(Mol · K). Measured value of a physical quantity is that value which is measured in localized lab or taking local physical conditions. The measured value may be equal to the accepted value or may be different to accepted value. Accepted value is represented by Accepted Value = xbest ± δx̄ Solved Problem 1.1 Two students measure the sound velocity at standard temperature and pressure (335 ± 4)m/s and (346 ± 2)m/s. Analyze their results and find out which experiment is more accurate and acceptable. Solution Two students measure the velocity of sound in air at standard temperature and pressure. Range of sound velocity as measured by first student is 335 − 4m/s ≤ vsf ≤ 335 + 4m/s 331m/s ≤ vsf ≤ 339m/s Similarly, range of sound velocity as measured by second student is 346 − 2m/s ≤ vsf ≤ 346 + 2m/s 344m/s ≤ vsf ≤ 348m/s Error data plot for both experiments done by two students is 0 1 2 330 332 334 336 338 340 342 344 346 348 vs O b b Accepted Value S1 S2

7. 1.1. MEASUREMENTS 7 It is seen that the range observe sound velocity by first student is lies near the accepted value of sound velocity. The discrepancy in observation of second student is 14m/s which is very large and his sound velocity range is not near the accepted value of sound velocity. This is why the observation by first student is reliable and acceptable. 1.1.5 Why Errors are Absolute? When we measured a quantity, it has certain value (c), that is perfectly measured by using scales and a residual/error value (e), that is approximated value rounded to least count of the measuring scale. This error may be subtractive or additive to the certain value. Therefore, a measured quantity is written in form of c ± e. See the following measurement. −1 0 1 2 3 4 5 6 7 8 9 10 Now, two students measure certain lengths of the thick wire. First student say, it is c1 = 5.1 and second student says, it is c2 = 5.2. This difference is due to difference in observations. The actual length of the thick wire is between 5.1 and 5.2. So, including error, that is equal to least count of the scale, i.e. one millimeter, first student will say that length of thick wire is 5.1 + 0.1 centimeter and second student will say that length of thick wire is 5.2 − 0.1 centimeter. But these two values can not be accepted at once. Therefore, we take result such that, student’s observations fall within its domain. So, we have length of thick wire either 5.1 ± 0.1 centimeter or 5.2 ± 0.1 centimeter. The domains of these two results are (i) 5.0 to 5.2 and (ii) 5.1 to 5.3. Students’ observations, i.e. 5.1 + 0.1 = 5.2 and 5.2 − 0.1 = 5.1 falls inside the domains of (i) and (ii). This is why, we takes error subtractive and additive. Now, come to the actual problem, Why Error is Absolute? Assume that length of two thick wires are −1 0 1 2 3 4 5 6 7 8 9 10

8. 8 Errors l1 = 5 ± 0.1 cm; l2 = 6 ± 0.1 cm The possible result values of l1 and l2 are 5−0.1 ≤ l1 ≤ 5+0.1 and 6−0.1 ≤ l1 ≤ 6+0.1 respectively. Sum of lengths is l = l1 +l2. The possible additions are l = 4.9 + 5.9 = 10.8 cm l = 4.9 + 6.1 = 11.0 cm l = 5.1 + 5.9 = 11.0 cm l = 5.1 + 6.1 = 11.2 cm These results shows that, the addition is 10.8 ≤ l ≤ 11.2, i.e. l = 11 ± 0.2. It means errors are added. In summary, l = (5 ± 0.1) + (6 ± 0.1) = (5 + 6) + (±0.1 + ±0.1) l = 11 + ±0.2 cm If there is difference of lengths of thick wires, i.e. l = l2 − l1. The possible additions are l = 5.9 − 4.9 = 1.0 cm l = 6.1 − 4.9 = 1.2 cm l = 5.9 − 5.1 = 0.8 cm l = 6.1 − 5.1 = 1.0 cm These results shows that, the subtraction is 0.8 ≤ l ≤ 1.2, i.e. l = 1.0 ± 0.2. Here, again errors are added. In summary, l = (6 ± 0.1) − (5 ± 0.1) = (6 − 5) + (±0.1 − ±0.1) l = 1.0 + ±0.2 cm This explains that, errors are absolute and they are always added to each other, irrespective of whether functions are either in addition, subtraction, multiplication or division.

9. 1.2. ERRORS 9 1.2 Errors In mathematics or in science or even in other streams of study, there are some fixed results of some problems/questions under certain conditions. For example, division of 12 by 4 always gives quotient 3. It shall not be 2.9999 or 3.1111 even though these values are mathematically in closed proximity of 3. During the observations, if result obtained by a student is not what, that is expected, then we say, that there is error of certain gravity. When there is error, there is different “output” other than desired “result”. Definition of error is given as Error is the deviation of practically measured quantity with its true (theoretical) quantity. Errors are defined by three types Systematic Error A student was asked to measure the length of card board with help of a scale. He measures the legth of card board several times very carefully and finds his observations as 22.5 centimenter. But actual length of card board was 22 centimeter. Now, why this variation exists even as student has measured the length of card board very carefully. It is due to the deformed scale. Measuring instruments are deformed due to their excessive use, wear & tear, elongations etc. Therefore, the error arises is called systematic error. This error generates due to malfunctioning of instruments, process and techniques etc. This is system error. Random Error As we explained above, measuring instruments gives different results under different environmental conditions. Physial properties of the instruments varies from season to season. For example, length of measuring scale changes from longest to shortest when environment changes from extreme summer to extreme cold. Barrometer shows different atmospheric pressure during summer and winter. This error happens without knowledge of the observer and it is called Random Error. This error depends on the physical properties of measuring instruments and measured item. This error happens randomly and can be avoid by stabilizing physical environment. Observational Error Precaution while observing an experiment is cru- cial for correct result of that experiment. A student, who is using digital speedometer to read speed of a car, observes a reading of 88.88 km per hour in place of 98.88 km per hour by mistake. Thus he makes an error of 10.00 km per hour. This is human error and mostly considered as observational errors. This error happens due to observations of observers. It can be avoided

10. 10 Errors by taking precautions. 1.2.1 Calculation of Errors Please be careful while you are handling errors. Errors are always errors and they can not be compensated by different observations. For example, a bank cashier of a bank performs two cash withdrawl transactions of Rs. 2000 each and handed over Rs. 1900/- and Rs. 2100/- respectively to two customers. Total money dispensed by him is Rs. 4000 and it is equally matched with financial book statement. But it can not be ignored that there is no error. The first customer who received less money shall submit an application regarding less dispense money and now it is banks liability to settle the matter by deducting Rs. 100 from the second cutomer and credit Rs. 100 to second customer and issuing notice to bank cashier. So, when a student measures two values as 99 and 101 in place of accurate result 100, then we can not say that the excess value in second observation may balanced to the less value of first observation. Error is always absolute. In first case measurement is less than the expected value, hence error is 99 − 100 = −1. Negative sign just tells that measured value is less than expected value. But actually error is | − 1| = 1. Similarly, in second case error is 101 − 100 = +1. Positive sign just tells that measured value is more than expected value. But actually error is | + 1| = 1. Total error is 1 + 1 = 2, i.e. errors are additive irrespective of their measured positive are negative values. Absolute Errors Absolute error is the deviation of measured quantity from mean of the all measured quantities. Let an instrument is used to measure the dia of a wire. Dia is measured ‘n’ times. The dia values measured by the instrument are a1, a2, a3, . . ., an respectively. Mean dia of the wire is amean = a1 + a2 + a3 + . . . + an n Error in the measured value of dia about mean is given by ai − amean. It may be negative or positive if mean error is larger than or lesser than the measured value. An error is error, either it is positive or negative. Therefore, absolute error is always taken as positive. To do so, we take the modulus of errors. So, Absolute error of first measurement is δa1 = |a1 − amean|

11. 1.2. ERRORS 11 Absolute error of second measurement is δa2 = |a2 − amean| Absolute error of third measurement is δa3 = |a3 − amean| Similarly, absolute errors of other measurements can be computed. Now absolute error of nth measurement is δan = |an − amean| Mean Absolute Error It is the mean of the absolute errors. For ‘n’ measurements mean absolute error is (δa)mean = |δa1| + |δa2| + |δa3| + | . . . | + |δan n Relative Absolute Mean Error It is the ratio of the mean absolute error of ‘n’ quantities to the mean of the measured ‘n’ quantities. It is given by (δa)mean amean . % Error Percentage Error is given by %e = (δa)mean amean × 100 Solved Problem 1.2 Find relative error of the fifth element measured from left hand side in the sampled data 85, 86, 84, 87, 88 and 83. Counting started from 1 to N. Solution The fifth element is 88. Mean of the all elements is m = 85 + 86 + 84 + 87 + 88 + 83 6 = 85.5 Absolute error of fifth element from mean is |88.0−85.5| = 2.5. Now relative error is E = 2.5 85.5 = 0.0292 This is absolute relative error which is equal to 2.92%.

12. 12 Errors Solved Problem 1.3 Find relative error of the fourth element measured from left hand side in the sampled data 55, 54, 56, 52, 53 and 57. Counting started from 0 to N. Solution When counting started from 0, 4th element is found at 5th place of an array when places are counted from 1. The fourth element in counting from 0 is 53. Mean of the all elements is m = 55 + 54 + 56 + 52 + 53 + 57 6 = 54.5 Absolute error of fourth element in counting from 0, about mean is |53.0 − 54.5| = 1.5. Now relative error is E = 1.5 54.5 = 0.0275 This is absolute relative error which is equal to 2.75%. Solved Problem 1.4 Find the % error of the sampled data 36, 36, 34, 35, 34, 32. Solution The mean of the sampled data is m = 36 + 36 + 34 + 35 + 34 + 32 6 = 34.5 The absolute errors of the sampled data from mean are δa1 = |36 − 34.5| = 1.5 δa2 = |36 − 34.5| = 1.5 δa3 = |34 − 34.5| = 0.5 δa4 = |35 − 34.5| = 0.5 δa5 = |34 − 34.5| = 0.5 δa6 = |32 − 34.5| = 2.5 Mean of the absolute errors is (δa)mean = 1.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 6 = 1.167 Percentage error of the sampled data about their mean is given by %e = (δa)mean amean × 100 = 1.167 34.5 × 100 = 3.38% This is percentage error of the sampled data from their mean.

13. 1.2. ERRORS 13 Root Mean Square Consider, δa1, δa2, . . ., δan are absolute errors of n successive observations about mean value amean. Now, Root Mean Square (rms) error is given by erms = r (δa1)2 + (δa2)2 + . . . + (δan)2 n Solved Problem 1.5 Find the RMS error of the sampled data 36, 36, 34, 35, 34 and 32 about their mean. Solution The mean of the sampled data is m = 36 + 36 + 34 + 35 + 34 + 32 6 = 34.5 The absolute errors of the sampled data from mean are δa1 = |36 − 34.5| = 1.5 δa2 = |36 − 34.5| = 1.5 δa3 = |34 − 34.5| = 0.5 δa4 = |35 − 34.5| = 0.5 δa5 = |34 − 34.5| = 0.5 δa6 = |32 − 34.5| = 2.5 Root Mean Square (rms) of the absolute errors is erms = r 1.52 + 1.52 + 0.52 + 0.52 + 0.52 + 2.52 6 Simplifying the right hand side, we have erms = r 2.25 + 2.25 + 0.25 + 0.25 + 0.25 + 6.25 6 = r 11.5 6 It gives, erms = √ 1.916 ≈ 1.38.

14. 14 Errors Solved Problem 1.6 Find the RMS error of the sampled data 85, 86, 84, 87, 88 and 83 about their mean. Solution The mean of the sampled data is m = 85 + 86 + 84 + 87 + 88 + 83 6 = 85.5 The absolute errors of the sampled data from mean are δa1 = |85 − 85.5| = 0.5 δa2 = |86 − 85.5| = 0.5 δa3 = |84 − 85.5| = 1.5 δa4 = |87 − 85.5| = 1.5 δa5 = |88 − 85.5| = 2.5 δa6 = |83 − 85.5| = 2.5 Root Mean Square (rms) of the absolute errors is erms = r 0.52 + 0.52 + 1.52 + 1.52 + 2.52 + 2.52 6 Simplifying the right hand side, we have erms = r 0.25 + 0.25 + 2.25 + 2.25 + 6.25 + 6.25 6 = r 17.5 6 It gives, erms = √ 2.917 ≈ 1.71. 1.2.2 Limitation of Errors An instrument which is used for observations, gives measurement errors due to any of the reasons, like design fault, wrong markers, environmental vari- ations and human errors. Though the measurement errors arise, but the maximum limit of error is limited by the least count of the instrument. For example, a meter scale that is used to measure the length of cloth or string has least count of 1cm, i.e. gap between two consecutive markers is 1cm. Similarly the students scale gives error upto 1mm. Vernier Calipers gives error upto 0.1mm while Screw Gauge gives error upto 0.01mm. Any computed

15. 1.2. ERRORS 15 error which is less than the least count of the instrument is rounded upward to the significant digit represent to least count of that instrument. Solved Problem 1.7 Mass of an object is measured multiple time to get the average mass of the object. The observed masses are 91kg, 98kg, 102kg and 101kg. If weighing machine can measured mass in multiple of 1kg, then find the average mass of the object. Solution Here the least count of the weighing machine is 1kg, i.e. machine can measure the mass of the object as multiple of 1kg. Hence any mean fraction shall be rounded off to whole number. Now the mean mass is m̄ = 91 + 98 + 102 + 101 4 = 98 The deviations of the measured values from the average value are δm1 = |(98 − 91)| = 7 δm2 = |(98 − 98)| = 0 δm3 = |(98 − 102)| = 4 and δm4 = |(98 − 101)| = 3 respectively. The average deviation is δm̄ = n P i=1 δmi n = 7 + 0 + 4 + 3 4 = 3.5 This deviation shall be rounded off to a factor of 1kg, i.e. 3.5 is rounded up to 4. Now the reported mean mass is 98 ± 4kg. 1.2.3 Error By Calculus Error in measurement depends on the relation between physical quantities. Assume a function of variable x (say) gives actual result at x = k. A student measures x and he found that x is not exactly equal to k but it is slightly deviates from k. He observes x = k ± δk. When this value of x is placed in the function, the function result also deviates from its actual result. This deviation is known as error. From the difference table of numerical calculus,

16. 16 Errors if change in the independent variable x is dx then corresponding change in dependent variable, y is given by dy. Here dy is called sample difference, and in measurement, it is called deviation of measured result from the actual result. dy of a function is obtained by derivating the given function about independent variable, here it is x. As mathematical operations are addition, subtraction, multiplication and division, hence there are four conditions in which errors are measured. Again, in any problem if error for a variable of a function is not given or is always zero, then that variable is considered as a constant (as Dk = 0). For Additive Quantities Let two quantities p and q are in summation, then y = p + q To find the maximum absolute error, we need to get the differentiation of the given relation: dy = dp + dq Here dy is error of quantity y measured and dp and dq are the errors in p and q quantities measured respectively. This shows that maximum error in addition is the arithmetic sum of errors in both quantities. For Subtraction Let two quantities p and q are in subtraction then y = p − q Now absolute error of measurement of quantity p is dp while absolute error of measurement of quantity q is dq then absolute error is dy = dp − dq This error would be maximum when dq is negative i.e. for maximum absolute error substituting dq = −dq, dy = dp + dq This is same as the maximum absolute error of additive measurement.

17. 1.2. ERRORS 17 For Product Quantities Let two quantities p and q are in product form as y = p × q Now differentiating this equation dy = p dq + q dp Dividing this relation by pq dy pq = dq q + dp p Substituting pq = y relation become dy y = dq q + dp p This is the maximum absolute error for productive quantities. For Quotient Quantities Let two quantities p and q are in fractional form as y = p q Now differentiating this equation dy = q dp − p dq q2 Dividing this relation by y dy y = q dp − p dq y q2 Substituting y = p q relation become dy y = dq q − dp p This error would be maximum when dq is negative i.e. for maximum absolute error substituting dq = −dq dy y = dq q + dp p This is the same for maximum absolute error for product of quantities.

18. 18 Errors Errors in Function of a Function Suppose f(s) = s is function of s and s is another function of time t as s(t) = t. The error involved in the measurement of the function f(s) is: d ds f(s) = d ds s = 1 Now, the absolute error is df(s) = ds. Here s is not independent as it is also a function of time t, hence we have to compute ds again. So, d dt s = d dt t It gives, ds = dt. Now the error in s can be replaced with error of t. So, the error shall be given by: df(s) = dt Solved Problem 1.8 A student measures the time period of pendulum by taking 100 observations in 90s. The least count of the clock used is 1s. If measurement of pendulum length is l = 20 ± 0.1cm, then find the % error in the measurement of gravitational constant. Solution The time period of the pendulum is given by T = 2π s l g The error in measurement of gravitational constant is obtained by derivating above relation. Assuming that, constants have no effect on the errors. Hence, ignoring the constants, squaring both sides and taking derivative of it w.r.t l. We have d dl T2 =

22. d dl l g

26. It gives 2T dT dl =

31. g d dl l − l d dl g g2

36. Or 2T dT = g g2 dl + l g2 dg

37. 1.2. ERRORS 19 Dividing left side by T2 and right side by its equivalent value l/g, we have 2 dT T = dl l + dg g The error in measurement of 100 oscillations is 1s. It is not for time period of the pendulum. 100 oscillations are completed by pendulum in 90s. It means, that the error of 1s rises in time measurement in 100 oscillations. Here, T is function of t. Again, we know that, T is given by t/100 as here t is total observation time for 100 oscillations. So, d dT T = d dT × t 100 On solving it we have dT = dt 100 ⇒ dT T = dt 100T = dt t Now, the error relation of pendulum becomes 2 × dt t = dl l + dg g Or 2 × dt t + dl l = dg g The negative sign of dl/l becomes positive as error is always absolute value. On substituting the known values, we have 2 × 1 90 + 0.1 20 = dg g = 0.0272 Now the percentage error in the measurement of g is dg g × 100 = 0.0272 × 100 = 2.72% This is required answer.

38. 20 Errors Solved Problem 1.9 What is percentage error in y for given relation y = x √ t ? We have given values t = 5 ± 0.1 and x = 1 ± 0.01. Solution To get the error in y, when there is error in x and t, we shall derivate y about x or t assuming than x and t both are independent variable of y. d dt y = d dt x √ t =

42. x −2t √ t

46. + 1 √ t dx dt The first term of right hand side is put within the two vertical lines (absolute symbol) to make error an absolute value as this term contains negative sign. Now dy = x dt 2t √ t + dx √ t Relative error in y is given by dy/y, so dy y = x dt 2t √ t x √ t + dx √ t x √ t Or dy y = dt 2t + dx x Percentage error in y is %y = dy y × 100 = dt 2t × 100 + dx x × 100 Substituting the values, we have %y = 0.1 2 × 5 × 100 + 0.01 1 × 100 It gives %y = 2%. Independent Variable Uncertainty Assume that q is a function of two independent and random variables x and y. x and y can be written as xbest ± δx and ybest ± δy respectively. Function q can be written as q = q(xbest, ybest) ± (|δqx| + |δqy|) (1.1)

47. 1.2. ERRORS 21 Uncertainty of the function q(x, y) is dq = (|δqx| + |δqy|) (1.2) Substituting the values of δqx and δqy in equation (1.2) dq ≈

51. ∂q ∂x δx

59. ∂q ∂y δy

63. (1.3) Uncertainty in a function of several variables Suppose that x, . . ., z are measured with uncertainties δx, . . ., δz and the measured values are used to compute the function q(x, . . . , z). If the uncertainties in x, . . ., z are independent and random, then the uncertainty in q is δq = s ∂q ∂x δx 2 + . . . + ∂q ∂z δz 2 (1.4) In any case it never larger than the ordinary sum δq ≤

67. ∂q ∂x

71. δx + . . . +

75. ∂q ∂z

79. δz (1.5) Solved Problem 1.10 Find the value of q = x3 y − xy3 where x and y are independent variables from each other. The given values are x = 4.0 ± 0.1 and y = 3.0 ± 0.1. Solution The values of x and y are xbest = 4.0, δx = 0.1 and ybest = 3.0, δy = 0.1 respectively. The best value of q is qbest = x3 bestybest − xbesty3 best That gives the value of q on substituting the best values of x and y. qbest = 64 × 3 − 4 × 27 = 192 − 108 = 84

80. 22 Errors Uncertainty in the q due to δx is δqx =

84. ∂q ∂x

88. δx = |3x2 y − y3 |δx = |144 − 27| × 0.1 = 11.7 Similarly, the uncertainty in q due to δy is δqy =

92. ∂q ∂y

96. δy = |x3 − 3xy2 |δy = |64 − 108| × 0.1 = 4.4 Finally, total uncertainty in q is the quadratic sum of these two partial uncertainties: δq = q (δqx)2 + (δqy)2 = p (11.7)2 + (4.4)2 = 12.5 Now the final result for q is q = 84 ± 12.5. Solved Problem 1.11 Find the error in function y = x2−1 x2+1 at x = 1 if dx = 0.1. Solution The given function is y = x2 − 1 x2 + 1 = 1 − 2 x2 + 1 Derivating it about x for finding of errors, we have dy dx =

101. 0 − (x2 + 1) × d dx 2 − 2 × d dx (x2 + 1) (x2 + 1)2 #

106. It gives dy dx = 4x (x2 + 1)2

107. 1.2. ERRORS 23 Substituting the values x = 1 and dx = 0.1, we have dy = 4 × 1 (12 + 1)2 × 0.1 = 0.1 This is desired result. 1.2.4 Application of Error Analysis Here are some problems that are solved to find the values with errors. These problems are sufficient to understand error problems. Solved Problem 1.12 Average age of a class is a function of a class, c in 0 c 5. The average age function is y = 8 + c2 /3, where y is in years. Find the average age of class c = 2 in years. Solution The average age of a class is given by function y = 8 + c2 3 The average age of the class c = 2 in years is given by y(c) and it is y(2) = 8 + 4 3 = 9.33 The average age of the class 2 is 9.33 years. Solved Problem 1.13 Find the acceptance value of the function f(s) = s+1/s when s = 2 ± 0.5. Solution The acceptance value of a function consists certain part and error part. s = 2 ± 0.5 has two parts. First certain part, i.e. s = 2 and second uncertain part, i.e. ds = 0.5. The certain value of the function is obtained at the certain point. So, f = 2 + 1 2 = 2.5 The error part of the function is df, when ds = 0.5. Derivating to the given function df ds =

111. d ds s

115. +

119. d ds 1 s

123. = 1 + 1 s2

124. 24 Errors Or df = 1 + 1 s2 ds Substituting the values, we have df = 1 + 1 4 × 0.5 = 0.625 Now the acceptance value is f ± df, or 2.5 ± 0.625. Solved Problem 1.14 In a physics problem, distance is observed as a function of time, i.e. s = t + t2 + sin t. Find the distance when t = 2s. Also, find the distance when t = 2 ± 0.1s. Solution Initially the distance is zero when t = 0. At time t = 2s, the distance is s = 2 + 22 + sin(2) Remeber that the standard argument of a trigonometric function is in radian. Solving above equation for distance s, the distance is 6.91 units. Student observes the time t = 2 ± 0.1s. Hence, the certain value of the function is obtained at t = 2s. While uncertain value is obtained at dt = 0.1s. Now, taking derivation of the distance function, we have ds = (|1| + |2t| + | cos t|)dt Using known values, we get ds = [|1| + |2 × 2| + | cos(2)|] × 0.1 = 0.54 The acceptance value of distance is 6.91 ± 0.54 units. Solved Problem 1.15 A student measures gravity by dropping a ball from a height of h and time t taken by it to reach at ground. After several observations he concluded that g = 49 ± 0.3m/s2 and t = 3.1 ± 0.06s. Find the error in the gravity. Solution The student measures gravity by applying height, time relation in vertical plane. The ball is dropped from the height of h with initial velocity u = 0 and it reached to ground in time t under the effect of gravity. Now the height is h = ut + 1 2 gt2 = 1 2 gt2 (1.6)

125. 1.2. ERRORS 25 Equation (1.6) gives the value of g g = 2h t2 The certain value of g is g = 2 × 49 9.8 g = 10 m s2 Now the relative error in the equation is given by ∂g g = ∂h h + 2 ∂t t (1.7) Here factor 2 has no discrepancy and can be eliminated from relation. The measurement of time is t = 3.1 ± 0.06s, this means certain value of time is 3.1s and there is error of ±0.06s. Hence relative error is ∂t t = 0.06 3.1 s (1.8) Similarly in measurement of height, there is certain value of height 49m and uncertainty 0.3m. Now relative error in the measurement of height is ∂h g = 0.3 49 m (1.9) Now the relative error in the gravity is ∂g g = 0.3 49 + 2 × 0.06 3.1 ∂g g = 0.0448 Now the error in the gravity is ∂g = 0.0448 × g (1.10) The error in gravity is ∂g = 10 × 0.0448 = 0.448m/s2

126. 26 Errors Now the final gravity is 10 ± 0.448 m s2 Ans. Solved Problem 1.16 Velocity of a bird in an interval of time t = 8.0 ± 0.1s are measured v1 = 0.21 ± 0.05 m/s and v2 = 0.85 ± 0.05 m/s respectively. Find the acceleration of the bird’s motion. Solution We know that the acceleration of an particle is aact = △vact △tact So actual acceleration is aact = 0.85 − 0.21 8 = 0.08 m s2 Now from error relation ∂a aact = ∂△v △vact + ∂t tact Where ∂ represents the error in physical quantity. Substituting the values of all variables ∂a 0.08 = 0.05 0.64 + 0.1 8 = 5 64 + 0.8 64 = 5.8 64 Now the error in acceleration measurement is ∂a = 5.8 64 × 0.08 = 7.25 × 10−3 m s2 Now measured acceleration is aact ± ∂a, So a = (80 ± 7.25) × 10−3 m s2 (1.11)

127. 1.2. ERRORS 27 The error graph is 0 1 2 70 72 74 76 78 80 82 84 86 88 a O b Actual Value A Solved Problem 1.17 Resistance of a conductor is given by ratio of the voltage across the conductor and current in the resistance. Find the error in measurement of the resistance if voltage and current is given as V = 2 ± 0.2 and i = 0.2 ± 0.02 Solution We know that the resistance of a conductor is given by Ohm’s law as Ract = Vact iact So actual resistance is Ract = 2 0.2 = 10Ω Now from error relation ∂R Ract = ∂V Vact + ∂i iact Where ∂ represents the error in physical quantity. Substituting the values of all variables ∂R 10 = 0.2 2 + 0.02 0.2 = 1 10 + 1 10 = 2 10 Now the error in resistance measurement is ∂R = 2 10 × 10 = 2Ω

128. 28 Errors Now measured resistance is Ract ± ∂R, So R = (10 ± 2)Ω (1.12) The error graph is 0 1 2 5 6 7 8 9 10 11 12 13 14 Ω b Actual Value A Solved Problem 1.18 Pressure of ideal gas is given by PV = nRT for n moles. Find the error in pressure measurement of gas if volume and temperature measurements are given by V = 12 ± 0.2lt and T = 289 ± 3K respectively. Solution For the ideal gas, pressure-volume relation is given by PV = nRT The certain pressure of the gas at V = 12lt and T = 289K is P = nR × 289 12 × 10−3 = 24.08nR × 103 Pa Now derivative of the ideal gas for pressure with respect to temperature dP dT =

133. nR × V − T × dV dT V 2

138. Or dP =

142. nR × V × dT − T × dV V 2

146. Substituting the values of variables, uncertain value is dP =

150. nR × 12 × 10−3 × 3 − 289 × 0.2 × 10−3 (12 × 10−3)2

154. Or dP = 0.123nR × 103 Pa

155. 1.2. ERRORS 29 Hence the pressure of the gas is P = (24.08 ± 0.123) nR × 103 Pa This is the pressure of the ideal gas when measured with possible errors. Solved Problem 1.19 A student measures two quantities m and a and obtains results m = 10.2 ± 0.2kg and a = 2 ± 0.2m/s2 . Now if he want to find product F = ma, find his answer. Also give both percentage and absolute uncertainties. Solution We know that the force on a particle of mass m is F = ma, where a is acceleration of moving particle. Fact = mact · aact So actual resistance is Fact = 10.2 × 2 = 20.4N Now from error relation ∂F Fact = ∂m mact + ∂a aact Substituting the values of all variables ∂F 20.4 = 0.2 10.2 + 0.2 2 = 1 51 + 1 10 = 6.1 51 Now the error in force measurement is ∂F = 6.1 51 × 20.4 = 2.44N Now measured force is Fact ± ∂F, So F = (20.4 ± 2.44)N (1.13) The error graph is

156. 30 Errors 0 1 15 16 17 18 19 20 21 22 23 24 N b Actual Value A Solved Problem 1.20 ‘x’ ‘y’ are measured as 2 ± 0.03 and 1 ± 0.07 respectively in radian during the angle measurement for the the relation f(x, y) = cos(2x + y2 ). find the value of the function f(x, y). Solution The given function is f(x, y) = cos(2x + y2 ) The certain value of the function is f(2, 1) = cos(2 × 2 + 12 ) = cos(5) = 0.284 Now the uncertain value of the function is obtained by derivation of it with respect to ‘x’. Hence df(x, y) =

158. − sin 2x + y2 [2 dx + 2y dy]

160. Substituting the values of the parameters, df(x, y) = |−0.2 × sin(5)| = 0.192 Now the measure value of the function is f(x, y) ≈ 0.284 ± 0.192 It is required result. Solved Problem 1.21 ‘x’ ‘y’ are measured as 30 ± 3 and 20 ± 2 respectively in radian during the angle measurement for the the relation f(x, y) = x 1 − cos y x . find the value of the function f(x, y). Solution The given function is f(x, y) = x h 1 − cos y x i

161. 1.2. ERRORS 31 The certain value of the function is f(30, 20) = 30 1 − cos 20 30 ≈ 6.42 Now the uncertain value of the function can be obtain by using derivative method like df(x, y) = h 1 − cos y x i |dx| + x

164. h 0 + sin y x i

167. ×

171. d dx y x

175. |dx| Or df(x, y) = h 1 − cos y x i dx + sin y x x dy + y dx x On substituting the parameter values df(30, 20) ≈ 3.11 Now the measured value is f(x, y) ≈ 6.42 ± 3.11 This is required result. Solved Problem 1.22 The radius of a sphere is measured as 2.1 ± 0.5cm. Find the surface area of the sphere with error limits. Solution Surface are of the sphere is given by A = 4πr2 Where r is the radius of the sphere. Certain measurement of the surface area of the sphere is A = 4π × 2.12 = 55.4cm2 Now the error in measurement of the surface area due to error in the measuring scale is dA = |8πr dr| On substituting the values of the parameters dA = 26.4cm2

176. 32 Errors Now the surface area measured with error limits is A = 55.4 ± 26.4 cm2 . Solved Problem 1.23 The voltage across a lamp is 6.0 ± 0.1volt and the current passing through it is 4.0 ± 0.2 ampere. Find the power consumed by the lamp. Solution Power consumed by an electrical instrument at potential V and current I is P = V I The certain power consumed by the instrument is P = 24 watt Now the error in measurement of the power consumed by the instrument is dP = |V dI + I dV | On substituting the values of the parameters dP = 1.6 watt Power consumed by instrument with error limits is P = 24.0 ± 1.6 watt. Solved Problem 1.24 The radius of curvature of a concave mirror measured by spherometer is given by R = l2 6h + h 2 The value of l and h are 4.0cm and 0.065cm respectively, where l is measured by a meter scale and h by a spherometer. Find the relative error in the measurement of R. Solution The radius of the curvature of the concave mirror is given by R = l2 6h + h 2 As, right hand side of the above relation is sum of two terms. as R = dP1 + dP2 Relative error of the relation is the sum of relative errors of the each term of the right hand side. Now dR R = dP1 P1 + dP2 P2

177. 1.2. ERRORS 33 First term is P1 = l2 6h Error in measurement of first term is dP1 = 1 6

181. 2lh dl − l2 dh h2

185. = 1 6 2lh dl + l2 dh h2 Now relative error of the first term of the relation is dP1 P1 = 2 dl l + dh h Similarly, relative error of the second term of the relation is dP2 P2 = dh h Now, dR R = 2 dl l + dh h + dh h Substituting the value of parameters dR R = 0.0807 This is the relative error may be arises during the measurement of the radius of the curvature. Solved Problem 1.25 The energy of a system as a function of time t is given as E(t) = A2 e−αt , where α = 0.2s−1 . The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50% then show that the percentage error in the value of E(t) at t = 5s is 4%. Solution Here both A and t has error of measurements. Now derivating the given function with respect to time. d dt E(t) = d dt A2 e−αt It gives dE(t) dt = A2 e−αt × −α + e−αt × 2A dA dt

186. 34 Errors Cross multiplication of dt dividing both side by E(t), above relation is re- duced to dE(t) E(t) = −α × dt + 2 × dA A Or dE(t) E(t) = −αt × dt t + 2 × dA A Multiplying both side by 100. dE(t) E(t) × 100 = −αt × dt t × 100 + 2 × dA A × 100 Now using the relation of % = △T T × 100 We have %E(t) = −αt × %et + 2 × %eA On substituting the values of time t = 5s and % errors of t A, we have %E(t) = 4% when t = 5s. Solved Problem 1.26 A pendulum is hanged in lab to compute the gravi- taional field intensity by a class 12 student. The length of simple pendulum is l. The time period of the pendulum at 25◦ C is 2s and given by T = 2π s l g If the temperature of the room changed to 26◦ C then find the change in the time period of the pendulum. The length of cord material of simple pendulum is a function of temperature t as lt = l0(1 + αt). Here, l0 is length of the chord at 0◦ C. Given, α = 1.25 × 10−2 /◦ C. Solution The time period of the pendulum is at temperature t◦ C is T = 2π s lt g Substituting the value of lt, we have T = 2π s l0(1 + αt) g

187. 1.2. ERRORS 35 Change in the time period, is obtained by derivating it in respect of t◦ C. So, d dt T = 2π s l0 g × d dt (1 + αt)1/2 Or dT dt = T × 1 2 (1 + αt)−1/2 α Expanding (1 + αt)−1/2 as binomial expansion and neglecting higher degree of terms of α. We have dT dt = Tα 2 × 1 − 1 2 αt dt = 26 − 25 = 1◦ C. Substituting the values, we have dT 1 = 2 × 1.25 × 10−2 2 × 1 − 1 2 × 1.25 × 10−2 × 25 It gives dT = 1.05 × 10−2 s. 1.2.5 Rounding Off Errors An instrument can measure the value upto its least count. Measured value less than least count of an instrument is known as the error of the measurement of that instrument. Any measured error that is less than its least count is rounded upward. For example, a meter scale can measure a length upto 1mm. Assume that length of a rod is calculated as 1235 ± 0.5mm. But for the meter scale, least count is 1mm. Here 0.5mm is the probable error of computation. But for the meter scale, least count is 1mm, hence this probable error is rounded upward to 1mm. Now for the meter scale, the computed value shall be 1235 ± 1mm. Solved Problem 1.27 A speedometer can measured the velocity of vehicle upto 1 meter per second correctly. A vehicle is moving with a speed v = t2 + t/2. At any instant of time t, velocity of the vehicle is measured. A watch is used to measure the time. The least count of the watch is 1s. Find the velocity of the vehicle at the time of t = 3s. Solution The velocity is function of time. Watch can measure the time upto 1s correctly. Hence maximum probability of measurement error in time

188. 36 Errors measurement is 1s. The least count of speedometer is 1m/s. Taking derivative of the velocity function with time t, dv dt = 2t + 1 2 Here dt = 1s, so its corresponding error in velocity measurement is dv = 2 × 3 + 1 2 × 1 = 6.5m/s Again, velocity of vehicle at t = 3s is vt = 32 + 3 2 = 10.5m/s The speedometer can measure the speed by 1m/s correctly, hence the measured speed of vehicle is v = 11 ± 7m/s. 1.2.6 Error in Sample Measurements When a physical quantity is measured multiple times, it gives different observations. Now, there is a confusion to select which observed value is correct. This confusion is eliminated by taking average of these observed values. The average of these measurements is again differ from the measured values. The difference between the measured values and their mean is called deviation of them. The deviations of the measured values from their mean is called as measurement errors. This is why, the result contains average value and deviation value both, like x̄ ± δx̄. The average of the measured values is obtained by using the mean relation as given below: x̄ = P |xn| n While the mean deviation is measured as given below: δx̄ = P |xn − x̄n| n Solved Problem 1.28 A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95s and 92s. If the minimum division in the measuring clock is 1s, then what shall be the reported mean time.

189. 1.2. ERRORS 37 Solution Here the least count of the clock is 1s. Hence any mean fraction shall be rounded off to whole number. Now the mean average is t̄ = 90 + 91 + 95 + 92 4 = 92 The deviation of the measured values from their average value are δt1 = |(90 − 92)| = 2 δt2 = |(91 − 92)| = 1 δt3 = |(95 − 92)| = 3 and δt4 = |(92 − 92)| = 0 respectively. The average deviation is δt̄ = n P i=1 δti n = 2 + 1 + 3 + 0 4 = 1.5 This deviation shall be rounded off to a multiple of 1s, i.e. 1.5 is rounded off upward to 2. Now the reported mean time is 92 ± 2 s. Solved Problem 1.29 Mass of an object is measured multiple time to get the average mass of the object. The observed masses are 91kg, 98kg, 102kg and 101kg. If weighing machine can measured mass in multiple of 0.5kg, then find the average mass of the object. Solution Here the least count of the weighing machine is 0.5kg, i.e. machine can measure the mass in multiple of 0.5kg. Therefore, mean fraction should be rounded off to the multiple of 0.5kg . Now the mean mass is m̄ = 91 + 98 + 102 + 101 4 = 98 The deviations of the measured values from the average value are δm1 = |(91 − 98)| = 7 δm2 = |(98 − 98)| = 0 δm3 = |(102 − 98)| = 4

190. 38 Errors and δm4 = |(101 − 98)| = 3 respectively. The average deviation is δm̄ = 4 P i=1 δmi 4 = 7 + 0 + 4 + 3 4 = 3.5 This mean deviation should be upward rounded off to a multiple of 0.5kg, i.e. 3.5 does not round off. Now the reported mean mass is 98 ± 3.5kg. Solved Problem 1.30 The diameter of a rod is measured with help of vernier caliper of least count 0.01cm. The observed diameters are 20, 21, 22, 24, 19, 19, 20 and 18 centimeters respectively. (a) What is the maximum absolute error of the measurements? (b) Find the mean error of observations. (c) What is mean rational error? (d) What is percentage error? Solution Here the least count of vernier caliper is 0.01cm, i.e. it can measure the diameter in the multiple of 0.01cm. Hence any mean fraction should be rounded off in the multiple of 0.01cm. Now the mean diameter is ¯ d = 20 + 21 + 22 + 24 + 19 + 19 + 20 + 18 8 = 20.375 The deviations of the measured diameters from the mean diameter are δd1 = |(20 − 20.375)| = 0.375 δd2 = |(21 − 20.375)| = 0.625 δd3 = |(22 − 20.375)| = 1.625 δd4 = |(24 − 20.375)| = 3.625 δd5 = |(19 − 20.375)| = 1.375 δd6 = |(19 − 20.375)| = 1.375 δd7 = |(20 − 20.375)| = 0.375 and δd8 = |(18 − 20.375)| = 2.375

191. 1.2. ERRORS 39 a The maximum absolute error of the measurements is found by upward rounding off of maximum absolute error (i.e. 3.625cm) in the multiple of 0.01cm, hence it is 3.63cm (upward rounding off). b The mean aabsolute error is δ ¯ d = 8 P i=1 δdi 8 Or δ ¯ d = 0.375 + 0.625 + 1.625 + 3.625 + 1.375 + 1.375 + 0.375 + 2.375 8 Or δ ¯ d = 1.46875cm This mean error should be upward rounded off to a multiple of 0.01cm. Now the mean error of observation is 1.47cm. c Mean Rational Error (MRE) is given by MRE = δ ¯ d ¯ d Here rational value is pure number without any unit, hence ratio shall be obtained by using upward rounded off values. So, now MRE = 1.47 20.38 = 0.07213 This is pure ratio without units, hence it does not need to be rounded off to a multiple of least count of vernier calliper. But we may apply rules of significant digits. Hence, MRE is 7.21 × 10−2 . d Percentage error is %e = δ ¯ d ¯ d × 100 = 1.47 20.38 × 100 = 7.213 This is also a number without unit. Hence, we may also apply rules of significant digits. So, percentage error is 7.21%.

192. 40 Errors 1.3 Definitions Accuracy The accuracy of an instrument is the extent to which the reading it gives might be wrong. Accuracy is often quoted as a percentage of the full-scale deflection (f.s.d.) of the instrument. For example, a one foot long school laboratory scale has full scale value of 12 inch with accuracy of ±1 millimeter. Repeatability The repeatability of an instrument is its ability to display the same reading for repeated applications of the same value of the quantity being measured. Reliability The reliability of an instrument is the probability that it will operate to an agreed level of performance under the conditions specified for its use. Reproducibility The reproducibility or stability of an instrument is its ability to display the same reading when it is used to measure a constant quantity over a period of time or when that quantity is measured on a number of occasions. Sensitivity The sensitivity of an instrument is given by ratio of the change in instrument scale reading to the change in the quantity being measured. In other words, the sensitivity of the instrument being measured is the ratio of the percentage change in the output quantity to the percentage change in the input quantity of the measuring instrument. Resolution The resolution or discrimination of an instrument is the smallest change in the quantity being measured that will produce an observable change in the reading of the instrument. The determination of the resolution of a measuring instrument is depends on the how finely smallest division (some time called least count, smallest scale division etc) is made on scale. Range The range of an instrument is the limits between which it can made readings. For example, range of one foot long scale of school lab is from one millimetre to twelve inch. Threshold The minimum quantity of measurement, that required before measuring instrument can response to it and able to give detectable reading, is called threshold.

193. 1.3. DEFINITIONS 41 Hysteresis Instruments can give different readings for the same value of measured quantity according to whether that value has been reached by a continuously increasing change or a continuously decreasing change. This effect is called hysteresis and it occurs as a result of such things as bearing friction and slack motion in gears in instruments. Error The error of a measurement is the difference between the result of the measurement and the true value of the quantity being measured. Instrument Error These errors arise from such causes as tolerances on the dimensions of components and electrical components used in the manufacture of an instrument. Insertion Error Insertion errors are errors which result from the insertion of the instrument into the position to measure a quantity affecting its value. For example, inserting an ammeter into a circuit to measure the current will change the value of the current due to the ammeter’s own resistance Environmental Error Environmental errors are errors which can arise as a result of environmental effects which are not taken account of, e.g., a change in temperature affecting the value of a resistance. Calibration Calibration can be defined as the process of determining the relationship between the values of the quantity being measured and what the instrument indicates. Calibration of an instrument is carried out by compar- ing the readings of instrument with the reading of the standard instruments for the same physical quantity. For example, one kilogram weighing mass is calibrated with the mass of the object placed in Peris for universal weight measurement. Primary Standards There are primary standards for mass, length, time, current, temperature and luminous intensity which are accepted by interna- tional agreements and are maintained by national establishments.

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