The length of a pendulum is (1.5 ± 0.01) m and the acceleration due to gravity is taken into account as (9.8 + 0.1) ms⁻². Calculate the time period of the pendulum with uncertainty in it. (2.5 ± 0.8 %) 


Given:

Length of the pendulum = l = (1.5 ± 0.01) m 
Acceleration due to gravity = g = (9.8 ± 0.1) ms⁻² 


To Find :

The time period of the pendulum = T = ?


Solution:

For division, percentage uncertainties are added. So, converting the fractional uncertainty to percentage uncertainty
l = (1.5 ± 0.01) m = (1.5 ± `\frac {0.01}{1.5}`ｘ100) m = (1.5 ± 0.667% ) m
g = (9.8 ± 0.1) m s⁻² = (9.8 ± `\frac {0.1}{9.8}`ｘ100) m s⁻² = (9.8 ± 1.02%) m s⁻²

Now by using the Time Period formula 

T = 2𝝿 `\sqrt frac {l}{g}`

T = 2𝝿 `\sqrt frac {(1.5 ± 0.687% ) m}{(9.8 ± 1.02%) m s⁻²}`

The fraction/percentage uncertainty is added in addition, subtraction, multiplication, and division. So,

T = 2𝝿 `\sqrt (frac {1.5}{9.8 s⁻²} ± (0.667% + 1.02% )) `

T = 2𝝿 `\sqrt frac {1.5}{9.8 s⁻²}`± `sqrt (0.667% + 1.02% ) `

T = 2𝝿 (0.39) s ±  `(1.687% ) ^(1/2)`

the power percentage uncertainty is multiply with the power

T = 2.458 s ±  `\frac {1}{2}`(1.69% )

T = 2.458 s ±  0.843 %

or

T = (2.458  ±  0.8 %) s   ------------Ans

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