Question

The period of oscillation of a simple
pendulum in the experiment is recorded
as 1.93 S, 1.94 S, 1.95 S, 1.93 S
respectively. The period with proper
error limits
(a) 1.93 + 0.01 s (b) 1.94 + 0.01 s
(c) 1.94 + 0.02 s (d) 1.93 + 0.02 s​

Answer 1

To Find :

• Absolute error

Solution :

period of oscillation of simple pendulum is recorded as 1.93 s, 1.94 s , 1.95 s ,1.93s

$\rm \: T = \frac{1.93 + 1.94 + 1.95 + 1.93}{4} \\ \\ \rm \: T = \frac{ 7.75}{4} \\ \\ \rm \: T =1.937$

T = 1.9

ABSOLUTE ERROR :

∆T1 = 1.93 - 1 .9 = +0.03

∆T2 = 1.94 - 1.9 = +0.04

∆T3 = 1.95 - 1.9 = +0.05

∆T4 = 1.93 - 1.9 = +0.03

Mean absolute error :

∆Tmean = (∆T1 + ∆T2 + ∆T3 + ∆T4)/4

= 0.03 + 0.04 + 0.05 + 0.03/4

= 0.15/4

= 0.037

Mean Absolute Error = 1.9±0.037

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