1. Measurement System1 Physical Measurement Method ( Metode Pengukuran Fisika) SF 091306 Gatut Yudoyono Physics department

2. Measurement Instrument2 No measurement can be made with perfect accuracy, but it is important to find out what the accuracy actually is and how different errors have entered into the measurement. Errors may come from different sources and are usually classified under three main headings: Gross Errors : Largely human errors, among them misreading of instruments, incorrect adjustment and improper application of instruments, and computational mistakes. Systematic Errors: Shortcomings of the instruments, such as defective or worn parts, and effects of the environment on the equipment or the user. Random Errors: Those due to causes that cannot be directly established because of random variations in the parameter or the system of measurement.

3. Gross Errors/ personal bias Measurement Instrument3 This class of errors mainly covers human mistakes in reading or using instruments and in recording and calculating measurement results. The position of pencil changes with respect to a mark on the background. Parallax error is introduced as we may read values at an angle. One common gross error, frequently committed by beginners in measure- ment work, involves the improper use of an instrument.

4. Measurement Instrument4 For example, a well-calibrated voltmeter may give a misleading reading when connected across two points in a high-resistance circuit. A voltmeter, having a sensitivity of 1,000  /V, reads 100 V on its 150-V scale when connected across an unknown resistor in series with a milliammeter. When the milliammeter reads 5 mA, calculate: a)apparent resistance of the unknown resistor, b)actual resistance of the unknown resistor, c)error due to the loading effect of the voltmeter. Solution a).The total circuit resistance equals Neglecting the resistance of the milliammeter, the value of the unknown resistor is R X = 20 k  b). The voltmeter resistance equals Since the voltmeter is in parallel with the unknown resistance, we can write c). Error

5. Measurement Instrument5 If the milliammeter reads 800 mA and the voltmeter reads 40 V on its 150-V scale. Solution a).The total circuit resistance equals Neglecting the resistance of the milliammeter, the value of the unknown resistor is R X = 50  b). The voltmeter resistance equals Since the voltmeter is in parallel with the unknown resistance, we can write c). Error

6. Measurement Instrument6 Systematic Errors This type of errors is usually divided into three different categories: 1.instrumental errors, defined as shortcomings of the instrument; 2. environmental errors, due to external conditions affecting the measurement. 3. Procedural error/measuring error

7. Measurement Instrument7 Instrumental errors are errors inherent in measuring instruments because of their mechanical structure. For example,  A zero error  In the d’Arsonval movement, friction in bearings of various moving components may cause incorrect readings.  Irregular spring tension, stretching of the spring, or reduction in tension due to improper handling or overloading of the instrument will result in errors  A calibration errors, causing the instrument to read high or low along its entire scale.

8. Measurement Instrument8 Instrumental errors may be avoided by: 1.selecting a suitable instrument for a particular measurement application; 2.applying correction factors after determining the amount of instrumental error; 3.calibrating the instrument against a standard. 4.replacing the instrument or by making a change in the design of the instrument.

9. Measurement Instrument9 Environmental errors are due to conditions external to the measuring device, including conditions in the area surrounding the instrument, such as the effects of changes in temperature, humidity, barometric pressure, or of magnetic or electrostatic fields. Thus a change in ambient temperature at which the instrument is used causes a change in the elastic properties of the spring in a moving-coil mechanism and so affects the reading of the instrument. Corrective measures to reduce these effects include air conditioning, hermetically sealing certain components in the instrument, use of magnetic shields, and the like.

10. Measurement Instrument10 Procedural error A faulty measuring process may include inappropriate physical environment, procedural mistakes and lack of understanding of the process of measurement. For example, if we are studying magnetic efect of current, then it would be erroneous to conduct the experiment in a place where strong currents are owing nearby. Similarly, while taking temperature of human body, it is important to know which of the human parts is more representative of body temperature. This error type can be minimized by periodic assessment of measurement process and improvising the system in consultation with subject expert or simply conducting an audit of the measuring process in the light of new facts and advancements.

11. Measurement Instrument11 Systematic errors can also be subdivided into static or dynamic errors.  Static errors are caused by limitations of the measuring device or the physical laws governing its behaviour. A static error is introduced in a micrometer when excessive pressure is applied in torquing the shaft.  Dynamic errors are caused by the instrument’s not responding fast enough to follow the changes in a measured variable.

12. Measurement Instrument12 Random Errors These errors are due to unknown causes and occur even when all systematic errors have been accounted for. In well-designed experiments, few random errors usually occur, but they become important in high-accuracy work. Suppose a voltage is being monitored by a voltmeter which is read at half-hour intervals. Although the instrument is operated under ideal environmental conditions and has been accurately calibrated before the measurement, it will be found that the readings vary slightly over the period of observation.

13. Measurement Instrument13 Random error unlike systematic error is not unidirectional.  Some of the measured values are greater than true value; some are less than true value.  The errors introduced are sometimes positive and sometimes negative with respect to true value. It is possible to minimize this type of error by repeating measurements and applying statistical technique to get closer value to the true value. No human being can repeat an action in exactly the same manner.

14. Measurement Instrument14 1. Least count error Least count error results due to the inadequacy of resolution of the instrument. The least count of a device is equal to the smallest division on the scale. Consider the meter scale that we use. What is its least count? Its smallest division is in millimeter (mm). Hence, its least count is 1 mm i.e. 10 -3 m i.e. 0.001 m. Clearly, this meter scale can be used to measure length from 10 -3 m to 1 m. It is worth to know that least count of a vernier scale is 10 -4 m and that of screw gauge and spherometer 10 -5 m.

15. Measurement Instrument15 Generally, the accepted level of error in reading the smallest division is considered half the least count. For example, let us read the measurement of a piece of a given rod. One end of the rod exactly matches with the zero of scale. Other end lies at the smallest markings at 0.477 m (= 47.7 cm = 477 mm). We may argue that measurement should be limited to the marking which can be denitely relied. If so, then we would report the length as 0.47 m, because we may not be denite about millimeter reading. Be more precise and accurate by reporting measurement as 0.477 ± some agreed level of anticipated error.

16. Measurement Instrument16 If we report the measurement in centimeter, If we report the measurement in millimeter,

17. Measurement Instrument17 2. Mean value of measurements It has been pointed out that random error, including that of least count error, can be minimized by repeating measurements. If we take average of the measurements from the repeated measurements, it is likely that we minimize error by canceling out errors in opposite directions. Here, we are implicitly assuming that measurement is free of systematic errors. The averaging of the repeated measurements, therefore, gives the best estimate of true value.

18. Measurement Instrument18 Significant Figures An indication of the precision of the measurement is obtained from the number of significant figures in which the result is expressed. Significant figures convey actual information regarding the magnitude and the measurement precision of a quantity. The more significant figures, the greater the precision of measurement. For example, if a resistor is specified as having a resistance of 68 , its resistance should be closer to 68  than to 67  or 69 . If the value of the resistor is described as 68.0 , it means that its resistance is closer to 68.0 than it is to 67.9  or 68.1 . In 68 there are two significant figures; in 68.0 there are three. The latter, with more significant figures, expresses a measurement of greater precision than the former.

19. Measurement Instrument19 Often, however, the total number of digits may not represent measurement precision. Frequently, large numbers with zeros before a decimal point are used for approximate populations or amounts of money. For example, the population of a city is reported in six figures as 380,000.  This may imply that the true value of the population lies between 379,999 and 380,001, which is six significant figures.  What is meant, however, is that the population is closer to 380,000 than 370,000 or 390,000. Since in this case the population can be reported only to two significant figures, how can large numbers be expressed?

20. Measurement Instrument20 A more technically correct notation uses powers of ten, 38 × 10 4 or 3.8 × 10 5. This indicates that the population figure is only accurate to two significant figures. Uncertainty caused by zeros to the left of the decimal point is therefore usually resolved by scientific notation using powers of ten. Reference to the velocity of light as 186,000 mi/s, for example, would cause no misunderstanding to anyone with a technical background. But 1.86 × 10 5 mi/s leaves no confusion.

21. Measurement Instrument21 Example 1 A set of independent voltage measurements taken by four observers was recorded as 117.02 V, 117.11 V, 117.08 V, and 117.03 V. Calculate 1. the average voltage, 2. the range of error. Solution 1. Average Voltage 2. Range Range = E max − E av = 117.11 − 117.06 = 0.05 V but also E av − E min = 117.06 − 117.02 = 0.04 V The average range of error therefore equals

22. Measurement Instrument22 Example 2 Two resistors R 1 and R 2, are connected in series. Individual resistance measurements, using a Wheatstone bridge, give R 1 = 18.7  and R 2 = 3.624 . Calculate the total resistance to the appropriate number of significant figures. Solution R 1 = 18.7  (three significant figures) R 2 = 3.624  (four significant figures) RT = R 1 + R 2 = 22.324  (five significant figures) = 22.3  The doubtful figures are written in italics to indicate that in the addition of R 1 and R 2 the last three digits of the sum are doubtful figures. When two or more measurements with different degrees of accuracy are added, the result is only as accurate as the least square measurement.

23. Measurement Instrument23 When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum. Example 3 In calculating voltage drop, a current of 3.18 A is recorded in a resistance of 35.68. Calculate the voltage drop across the resistor to the appropriate number of significant figures. Solution E = IR = 3.18 × 35.68 = 113.4624 = 113V Since there are three significant figures involved in the multiplication, the answer can be written only to a maximum of three significant figures.

24. Measurement Instrument24 Example 4 Add 826 ± 5 to 628 ± 3 Solution N 1 = 826 ± 5 (= ±0.605%) N 2 = 628 ± 3 (= ±0.477%) Sum = 1454 ± 8 (= ±0.55%) Note in Example 4 that the doubtful parts are added, since the ± sign means that one number may be high and the other low. The worst possible combination of range of doubt should be taken in the answer. The percentage doubt in the original figure N 1 and N 2 does not differ greatly from the percentage doubt in the final result.

25. Measurement Instrument25 Example 5 Subtract 628 ± 3 from 826 ± 5 and express the range of doubt in the answer as a percentage. Solution N 1 = 826 ± 5 (= ±0.605%) N 2 = 628 ± 3 (= ±0.477%) Difference = 198 ± 8 (= ±4.04%) Again in Example 5, the doubtful parts are added for the same reason as in Example 4. Comparing the results of addition and subtraction of the same numbers in Example 4 and 5, note that the precision of the results, when expressed in percentages, differs greatly. The final result after subtraction shows a large increase in percentage doubt compared to the percentage doubt after addition. The percentage doubt increases even more when the difference between the numbers is relatively small.

26. Measurement Instrument26 Example 6 Subtract 437 ± 4 from 462 ± 4 and express the range of doubt in the answer as a percentage. Solution N 1 = 462 ± 4 (= ±0.87%) N 2 = 437 ± 4 (= ±0.92%) Difference = 25 ± 8 (= ±32%)