We measure the period of oscillation of a simple pendulum. In successive Measurements the readings turn out to be 2.63 S, 2.56 S, 2.42 S, 2.71 S, and 2.80 S. Calculate the absolute errors, relative error and percentage error
Answers
Answer:
Readings,
x
1
=
2.63
s
x
2
=
2.56
s
x
3
=
2.42
s
x
4
=
2.71
s
x
5
=
2.80
s
First calculate the arithmetic mean of readings,
x
m
e
a
n
=
x
1
+
x
2
+
x
3
+
x
4
+
x
5
5
x
m
e
a
n
=
2.63
+
2.56
+
2.42
+
2.71
+
2.80
5
x
m
e
a
n
=
2.62
s
The absolute error can be calculated as,
For first reading,
Δ
x
1
=
|
2.63
−
2.62
|
Δ
x
1
=
0.01
s
For second reading,
Δ
x
2
=
|
2.56
−
2.62
|
Δ
x
2
=
0.06
s
For third reading,
Δ
x
3
=
|
2.42
−
2.62
|
Δ
x
3
=
0.20
s
For forth reading,
Δ
x
4
=
|
2.71
−
2.62
|
Δ
x
4
=
0.09
s
For fifth reading,
Δ
x
5
=
|
2.80
−
2.62
|
Δ
x
5
=
0.18
s
Calculate the arithmetic mean of absolute errors,
Δ
x
m
e
a
n
=
Δ
x
1
+
Δ
x
2
+
Δ
x
3
+
Δ
x
4
+
Δ
x
5
5
Δ
x
m
e
a
n
=
0.01
+
0.06
+
0.20
+
0.09
+
0.18
5
Δ
x
m
e
a
n
=
0.54
5
Δ
x
m
e
a
n
=
0.11
s
Calculate the relative error,
R
E
=
Δ
x
m
e
a
n
x
m
e
a
n
R
E
=
0.11
2.62
R
E
=
0.04
s
Calculate the percentage error,
P
E
=
Δ
x
m
e
a
n
x
m
e
a
n
×
100
P
E
=
0.11
2.62
×
100
P
E
=
4
%
Arithmetic mean,a mean = 2.63+2.56+2.42+2.71+2.80 / 5
= 13.12 / 5
a mean = 2.624 s
Absolute error,
delta a1 = 2.63 - 2.62
= 0.01 s
delta a2 = 2.56 - 2.62
= 0.06 s
delta a3 = 2.42 - 2.62
=0.20 s
delta a4 = 2.71 - 2.62
= 0.09 s
delta a5 = 2.80 - 2.62
= 0.18 s
delta a mean =|delta a1|+|delta a2|+|delta a3|+|delta a4|+|delta a5| / 5
= 0.108
= 0.11 s
a = a mean (+/-) delta a mean
= 2.62 (+/-) 0.11
=2.51 < 2.62 < 2.73
Relative error = 0.11 / 2.6
= 0.04 s
Percentage error = delta a mean / a mean x 100
= 0.11 / 2.6 x 100
= 4 %