In an experiment of simple pendulum, a student made several observations for the period of oscillation. His readings turned out to be 2.63s , 2.56s , 2.42s , 2.71s and 2.80s. Find :-
1) Mean Period of oscillation or most accurate value of Time period.
2) Absolute error in each reading.
3) Mean absolute error.
4) Fractional and Percentage error.
Answers
The mean period of oscillation of the pendulum is
T
mean
=
n
∑
i=1
n
T
i
;T
mean
=
5
(2.63+2.56+2.42+2.71+2.80)
s
5
13.12
s=2.624s=2.62s
(Rounded off to two decimal places)
The absolute errors in the measurement are
ΔT
1
=2.62s−2.63s=0.01s;ΔT
2
=2.62s−2.56s=0.06s
ΔT
3
=2.62s−2.42s=0.20s;ΔT
4
=−2.71s−0.09s
ΔT
5
=2.62s−2.80s=−0.18s
Mean absolute error is
T
mean
=
n
∑
i=1
n
∣ΔT
i
∣
T
mean
=
5
0.01+0.06+0.20+0.09+0.18
s
=
5
0.54
s=0.11s
Answer:
(i) The mean time period of oscillation
T=2.63+2.56+2.42+2.71+2.8052.63+2.56+2.42+2.71+2.805
=13.125s=2.624s≈2.62s=13.125s=2.624s≈2.62s
(rounded off to 2nd decimal palce )
(ii) Taking 2.62s as the true value the absolute errors (true value - measured value ) in the five readings are,
(2.62 - 2.63) s= -0.01s, (2.62-2.56) s = 0.06s,
(2.62 - 2.42)s = 0.20s, (2.62 - 2.71)s = -0.09 s and
(2.62-2.80) s =-0.18s
(iii) The (maximum) mean absolute error is
(δT)max=0.01+0.06+0.20+0.09+0.185(δT)max=0.01+0.06+0.20+0.09+0.185s
=0.545s=0.108s≈0.11=0.545s=0.108s≈0.11s
(iv) The (maximum) fractional error is
(δTT)max=0.11s2.62s≈0.04(δTT)max=0.11s2.62s≈0.04
(v) The maximum percentage error is
(δTT)max×100=0.04×100=4%(δTT)max×100=0.04×100=4%
∴∴ The value of T should be written as (2.62±0.11)2.62±0.11)s
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