The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s,2.56s,2.42s,2.71s, and 2.80s respectively. find the time period
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2
Answer:
0.11 sec
EXPLANATION:
Correct option is
B
0.11s
T=
5
2.63+2.56+2.42+2.71+2.80
5
13.12
=2.624=2.62
2.62−2.63=−0.01
2.62−2.56=+0.06
2.62−2.42=+0.20
2.62−2.71=−0.09
2.62−2.80=−0.18
Average absolute error
5
0.01+0.06+0.20+0.09+0.18
0.545=0.108=0.11s
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