physics 1 lab -propagation of uncertainties

topic 1 : propagation of uncertainties
uncertainty : intrinsic er ror i n estimating measurements question 2a
·
·




54 cm
·

question I f dividing uncertainties propagation
I I De = uncertainty
·

convert value (4 . 9170 + 0 . 0005) cm to inches ,
1 inch = 2 .




+War =A
br
↑
. 9170cm
.Finch
4 =


central value = 1 .
as r



e
er ror propagation ↳ for r
·
=




Linch
0 . 0005cm .


W
0 0001969 . 0
0
your measurements affect the results derived from measurements
= =
how the uncertainty in
.
·




the uncertainties propagate into final results
(0 .005000S 23 . .
·




ar = .




(1 9358 = 0 .
.
0002) in
·

er ror due to limitation of equipment
(1 . 222 + 0 006)
.




·

determine uncertainty by smallest value device can measure ·



question 2b
eg : a ruler whose smallest value is I mm will have an uncertainty of 10 S of
.
mm ·
what is the value
= + dz =
dd its
uncertainty ? can make multiple measurements a their results
·
we average
eg : measured a value of 4 95 cm on ruler whose smallest value is 0 . 1 ,
cm uncertainty is 10 05 cm d1 =
(1 000 + 0 005)
our estimate would be the
. .




value
.




mean
.
·




(4 95 = 0 05) cm (0
d2 =
230 = 0 005)
=
.
.
.
.




e
li * a smaller mean more accurately represents the data
·



digital caliper measures to nearest 100th of a mm 0 . 01 mm
minimum : i


eg : caliper measures 49 17mm
, uncertainty = = 0 005 mm
we can calculate the estimate of this uncertainty via standard deviation
.




1. 000 005 995
.




0
·

-

0 .
= .




D1 22
if
↳ you conver t to cm : 4 9170 cm uncertainty = 10 0005cm .




0 230 -0 005
. .

,
.
.
= 0 225 li = value
Or
.




Ill
·

question I a
maximum :
l = mean

·

read 49 17 mm but fluctuates btwn
. 49 Olmm 49 22mm what is the error ? 1 000 + 0 005 = 1 005
Al 6 l
.
,
.
. .
.
,


D 1 .
24
=




difference btwn lowest value : 0 .
230 + 0 .
005 =
0 .
235

the larger the standard deviation , the more spread out the mean is
49 .
17mm-49 .
Olmm =
0 16 mm
.




average :

difference btwn highest value :
1 .
22 + 1 . 24
= 1 23
05
.




49 .
22mm-49 .
17mm = 0 . mm
2

video practice problems
average error : :
uncertainty

D = AXmin + DX max
0 11 mm 0 0005 + 0 0005 = 0 001
equation : a =
xy +
4x(5y +
5) ,
what is Da ?
0 105mm
.

=
.
.


= .
.




2 in order to find Ad follow PEMDAS
,
~
according to rounding : 0 0010
.




error :
e r ro r :
.
1 start w/ by +
24 treat each term independent i

(49 17 = 0 11) mm

-)
.




010)
.




(1 . 230 + 0 . m
A5y292b(sy +




8-value . next
2 ,
multiplication rule w/4 x


·
if 8 < 1 , the two values agree
·

question 2C adding uncer tainties propagation
·


errors might be too large or overestimated given X A X & Y 1Dy , determine D2 in ter ms of DX & Dy
.
b(4x(Sy +
-
·
if 1 < f < 3 ,
the two values are intension 2 = x +
y . find
3 uncertainty of Xy

D2 =
DX +
DY
·
if 8 < 3 the two values do not agree DXY
,



errors might be too small 4. uncertainty of combining all terms
question 2d
·




subtracting
a is
uncertainties propagation
·




equation : given X A X & Y 1Dy determine Aw in ter ms of DX & Dy
.
-
·

,
Ad =
x(xy + 4x(5y +



Ile-12) w = X Y
& value must be rounded to
-




Se *


same amt. of sigtigs as the value
(Dl1)2 + (De2)
<
AW = Dx +
AY 1
.
uncertainty of F = ma
wl the lowest amt. of sigfigs
a
.




question ze multiplying uncertainties propagation
·




·


rounding uncertainties multiplication rule :

Be Be AWw
DA
D
It would result a value of 1
= ,
·
round uncertainty to one significant figure unless in
A
D DA =

x(xy) = DX -



y +
XDY
W
·

then round your central value to the same decimal place as your rounded uncertainty
DF =
Am .


a + m Da
for 1
.




↳ A = w
eg : (4 225 + 0 0398) .
. p (4 .
23 = 0 . 04)

= 0 1293) (12 09 + 0 13)
eg : (12
093
.
.
a .
.


1 = (2 .
310 1 0 .
005) m w = (1 890 ! 0 005)m
. .




A =

(0 .0050005 .


(2 310
.
-


1890 0 =




(4 37 10 02)m2 . .