tq could you include information on the calculation of uncertainties and the uncertainties in using different measurements? you posted the relevant information and removed it so I was wondering if you could post it again. thanks
here u go
random errors:
* Unpredictable changes in measurements
* can be effected by the environment
* nearly always human [parallax] error
* can be reduced by repetition and averages
* correct readings will always be between the limits after repeated experiments
* use better instruments
* scatter results about value.
Systematic errors:
* same change between measurements (+/-2)
* Affected by flaws in instruments
* human error only when instruments wrongly used.
* cannot be reduced without changes in instruments (recalibration etc)
* accuracy reduced
* results in all readings being too large or too small
Uncertainties in some instruments:
* Stopwatch: 0.2 secs
* Metre rule: 1mm
* Vernier caliper: 0.1mm
* Micrometer: 0.01mm
You want to measure the length of a wire. The reading is between 374mm and 375mm. You judge it is closer to 374mm. Since you can only judge it to the nearest mm (the smallest on the ruler). You write: l
length of wire: 375mm +/- 1mm
To reduce the error, use a different instrument.
Error is shown by number of sig figs and the +/- signs.
eg. A voltmeter might read: 1.25v +/-0.01v. This implies voltage between 1.24 and 1.26
Actual uncertainty therefore would be:
1mm for the rule
0.1 for the voltmeter
0.01mm for the micrometer
This can also be written as % uncertainty using the following formula:
uncertainty in the instrument/reading *100
1/375*100= +/-0.26%
Combining uncertainty:
In some cases we have to combine the uncertainties:
1. Adding or subtracting uncertainty: add together their absolute uncertainty to obtain the absolute uncertainty.
2. Multiply or dividing uncertainty: add together their % uncertainty to obtain % uncertainty
3. Raised to the power n: Multiply % uncertainty by n to obtain the % uncertainty.
Examples:
1. Finding R using a voltmeter and an ammeter if the reading on the voltmeter is 2.45v and the ammeter is 0.96A:
V=2.54 +/-0.01v
% error = 0.01/2.54*100 = 4%
I= 0.96A +/-A
0.01/0.96*100= 1%
combined error: 4%+1%= 5%
r=v/i
2.54/0.96 =2.56 +/- 5% ohms
2.% uncertainty in multiple readings: range/2*100
eg. 1.96mm, 1.94mm, 1.98mm, 1.95mm, 1.97mm
(1.98-1.94)/2*100 = 2%
3. % difference:
difference/average*100
Used to compare your answer to examiner's or manufacturer's.
You find the density of plasticine to be 1.8g/cm3
The manufacturer gives the density as 1.6g/cm3
The % difference: 1.8-1.6/[(1.8+1.6)/2] * 100= 11.8%
If you found the % uncertainty= 5%
5% of 1.8= 0.1
your density= 1.8+/- 0.1
your density is between 1.7 and 1.9 g/cm3
The manufacturer's density is 1.6g/cm3 which is outside your range.
Therefore, it is a different plasticine.
