Question

The length of a simple pendulum is about $ 100cm $ , known to have an accuracy of $ 1mm $ . Its period of oscillation is $ 2s $ , determined by measuring the time for $ 100 $ oscillations using a clock of $ 0.1s $ resolution. What is the accuracy in the determined value of $ g $ ?
(A) $ 0.2\% $
(B) $ 0.5\% $
(C) $ 0.1\% $
(D) $ 2\% $

Answer 1

Hint
To solve this problem we can use the formula of the time period of a pendulum to find the equation in terms of $ g $ and then using the resolution of the clock we can find the $ g $ . From there we have to find $ \% $ accuracy.
$\Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}} $
where, $ T $ is the time period of oscillation of the pendulum,
$\Rightarrow l $ is the length of the pendulum,
$\Rightarrow g $ is the acceleration due to gravity.
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{\Delta l}}{l} + 2\dfrac{{\Delta T}}{T} $
 where $ \Delta g $ is the small change in acceleration due to gravity,
$\Rightarrow \Delta l $ is the small change in length of the pendulum,
$\Rightarrow \Delta T $ is the small change in time period.
And $ \Delta T = \dfrac{{{\text{Resolution of the clock}}}}{{{\text{Number of oscillations}}}} $ .

Complete step by step answer
The formula to calculate the time period of oscillation of a pendulum is,
$\Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}} $
From here we can convert the equation in terms of $ g $ since we need to find the accuracy of $ g $ .
So, we square both the sides and then take $ g $ in the R.H.S to the L.H S and $ {T^2} $ in the L.H.S to the R.H.S.
$\therefore g = {\left( {\dfrac{{2\pi }}{T}} \right)^2} \times l $
The accuracy $ \% $ of $ g $ can be found out by dividing a small change in the value of acceleration due to gravity $ \Delta g $ by the value of $ g $ and multiplying it by $ 100\% $ .
$\Rightarrow \dfrac{{\Delta g}}{g} \times 100\% $
So, for the $ \% $ accuracy, we need to use the smallest change in the other variables $ T $ and $ l $ .
$\therefore \dfrac{{\Delta g}}{g} = \dfrac{{\Delta l}}{l} + 2\dfrac{{\Delta T}}{T} $
We have multiplied 2 by $ \dfrac{{\Delta T}}{T} $ because the term $ T $ is squared in the formula of $ g $ .
So, to find the percentage accuracy, we multiply by $ 100\% $ on both the sides of the equation. $ \therefore \dfrac{{\Delta g}}{g} \times 100\% = \dfrac{{\Delta l}}{l} \times 100\% + 2\dfrac{{\Delta T}}{T} \times 100\% $
Now, from the question,
$\Rightarrow \Delta l = 1mm = 0.1cm $
$\Rightarrow l = 100cm $
and $ \Delta T = \dfrac{{{\text{Resolution of the clock}}}}{{{\text{Number of oscillations}}}} = \dfrac{{0.1}}{{100}} = 0.001s $
$\Rightarrow T = 2s $
So, substituting the values in the equation, we get
$\Rightarrow \dfrac{{\Delta g}}{g} \times 100\% = \dfrac{{0.1}}{{100}} \times 100\% + 2\dfrac{{0.001}}{2} \times 100\% $
$\Rightarrow \dfrac{{\Delta g}}{g} \times 100\% = 0.1\% + 0.1\% = 0.2\% $
Thus we find the accuracy of $ g $ to be $ 0.2\% $ .
So, the correct option is (A).

Additional Information
The time period of a simple pendulum doesn’t depend on the mass of the bob or the value of the initial angular displacement, but rather depends on the length and the acceleration due to gravity.

Note
The accuracy in the value of $ g $ can be increased easily by measuring the length of the string of the pendulum with the help of an instrument which has the smallest division even less than $ 1mm $ and while measuring the time, by increasing the number of oscillations. Taking the value of more number of oscillations decreases the chances of making errors.