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Lab Report 1 of Applied Physics. Topics covered are: Error Analysis. All the necessary calculations are done.
Typology: Study Guides, Projects, Research
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Abstract:
theoretical distribution of a large population of such
estimates.
Standard error=
σ ´ x
s
√ n
Errors and their Types:
Error:
Error is defined as the difference between a
measured value and the expected value.
Types of Errors:
There are two types of errors:
Random error
Systematic error
Random error:
Random errors are statistical fluctuations in the
measured data due to the precision limitations of
the measurement device.Random errors can be
evaluated by statistical analysis and can be reduced
by averaging over a large number of obsrevations.
Systematic error:
Systematic errors are reproducible inaccuracies
that are consistently in the same direction.These
errors are difficult to detect and cannot be analyzed
by statistically.If a systematic error is identified
when calibrating against a standard,applying a
correction or a correction factor to compensate for
the effect can reduce the bias.Unlike random
errors,systematic errors cannot be detected or
reduced by increasing the number of observations.
Standard deviation in length = (S.D)
=
√
2
Standard deviation in width = (S.D)
=
√
2
5. We calculated the standard error in length and
width by using the following formulas:
Standard deviation in length= (S.E)
=
√
L
2
Standard deviation in width= (S.E)
=
√
W
2
6. We calculated the standard error in
perimeter[(S.E)
] and area[(S.E)
] of the register by
using the following formulas:
(S.E)
=
√
(
)
2
L
2
(
)
2
W
2
(S.E)
=
√
(
)
2
L
2
(
)
2
W
2
Calculations:
Length
(cm)
Width
(cm)
Deviation
in length
o
Deviation
in width
o
(cm) (cm)
27.3 21.6 -0.08 -0.
27.4 21.7 0.02 0.
27.3 21.8 -0.08 0.
27.5 21.6 0.12 -0.
L
√
2
√
2
+(0.02)
2
+(−0.08)
2
+(0.12)
2
¿ √0.
¿ 0.096 cm
W
√
2
+(0.02)
2
+(0.12)
2
+(−0.08)
2
¿ (^) √0.0092=0.096 cm
L
√
L
2
√
2
=√0.
¿ 0.048 cm
W
√
W
2
√
2
=√0.
¿ 0.048 cm
Standard error in Parameter:
Experiment No.
Error analysis technique for finding the value
of “g” using a simple pendulum.
Procedure:
1. We measured the length of the string from it’s
one end to the middle of the metal bob.We took 3
readings and calculated the mean length
.
2. Now,we attached the string to an iron stand and
measured the time ‘t’ for 10 vibrations/oscillations
of the simple pendulum.Then,we calculated the
time period ‘T’ for one oscillation by:
t
We calculated the mean time
.
3. We calculated the deviation in length:
o
And deviation in time:
o
4. We calculated the standard deviation in length
and time by using the following formulas:
Standard deviation in length= (S.D)
=
√
2
Standard deviation in time= (S.D)
=
√
2
(s) (cm) (s)
44.0 14.34 1.434 0.1 0.
43.9 12.57 1.257 0 -0.
43.8 13.69 1.369 -0.1 0.
L
√
2
√
2
+( 0 )
2
2
√
=√0.01=0.
L
=0.1 cm
T
√
2
√
2
+(−0.096 )
2
2
¿ √0.
¿ 0.282 s
L
√
L
2
=√0.003=0.0578 cm
T
√
T
2
=√0.026=0.163 s
∂ g
(
4 π
2
l
2 )
= 4 π
2
l T
− 3
(− 2 ) =− 8 π
2
l T
− 3
¿− 8 π
2
( (^) 43.9) ( (^) 1.35) (^) =1399.
∂ g
4 π
2
2
4 π
2
2
g
√
L
2
(
∂ g
)
2
T
2
(
∂ g
)
2
¿ (^) √( 0.058)
2
2
2
2
¿ (^) √50,
¿ 223.84 cm s
− 2
g =
4 π
2
l
2
g
g =(^ 946.58 ± 223.84 )^ cm s
− 2
Percentage error in ‘g’’:
% error =
(
)
The negative sign indicates that the measured value
is less than the actual value.
Conclusion:
The value of ‘g’ calculated by simple pendulum is:
g =(^ 946.58 ± 223.84 )^ cm s
− 2
