Question

the time period of oscillation of a simple pendulum in an experiment is recorded as 2.56s, 2.62s, 2.70s, 2.58s, 2.45s respectively. Find the time period, mean absolute error and percentage error.​

Answer 1

Explanation:

A1=2.56

A2=2.62

A3=2.70

A4=2.58

A5=2.45

avg a=2.582

delta a=absolute error in the measurements are=(2.56-2.58)=-0.02

(2.62-2.62)=0

(2.70-2.62)=0.08

(2.58-2.62)=-0.04

(2.45-2.62)=-0.17

mean absolute errors=(-0.02+0+0.08-0.04-0.17)/5=-0.03

relative error=±0.03/2.58=±0.011

percentage error=1.16%

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Answer 2

Answer:

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