Question

The length of a simple pendulum is about \(100~cm\) known to an accuracy of \(2~mm\). Its period of oscillation is\( 2~s\) determined by measuring the time for \(500\) oscillations using a clock of \(1~s\) resolution. What is the accuracy (in percentage) in the determined value of \(g\) ?

Answer 1

Find accuracy in period of oscillation.

Given, period of oscillation, \(T=2~s\)

And \(500\) oscillations using a clock of \(1~s\) resolution,

So, accuracy in period of oscillation,

\(\Delta T= \dfrac {1}{500}=0.002~s\)

Find accuracy in value of \(g\).

Given, length of simple pendulum, \( l=100~cm\)

Accuracy in length, \(\Delta l=2~mm=0.2~cm\)

As we know, time period of simple pendulum

\(T = 2 \pi \sqrt{\dfrac{l}{g}}\)

\(g = \dfrac{4 \pi^2 l}{T^2}\)

Differentiating the above equation,

\(\dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + \dfrac{2 \Delta T}{T}\)

\(\dfrac{\Delta g}{g} = \dfrac{0.2}{100} + 2\left
(\dfrac{0.002}{2}\right )\)

Find percentage error in value of \(g\) .

Percentage error in value of \(g\).

\(\dfrac{\Delta g}{g} \times 100 = \dfrac{0.2}{100} \times 100 + 2\left ( \dfrac{0.002}{2} \right ) \times 100\)

\(\dfrac{\Delta g}{g} \times 100 = 0.2 + 0.2\)

\(\dfrac{\Delta g}{g} \times 100 = 0.4\)

Final answer : \(0.40\)