First slide

Accuracy; precision of instruments and errors in measurements

Question

A simple pendulum is being used to determine the value of acceleration due to gravity ‘g’ at a certain place. The length of the pendulum is measured as 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50s. The accuracy in ‘g’ is :

Moderate

A

2.40%
B

5.40%
C

3.40%
D

4.40%

Solution

$\begin{array}{l} T i m e p e r i o d = T = 2 π \sqrt{\frac{L}{g}} \\ \Rightarrow g = 4 π^{2} \frac{L}{T^{2}} \\ \Rightarrow \frac{Δ g}{g} = \frac{2 Δ T}{T} + \frac{Δ L}{L} = 2 (\frac{1}{50}) + \frac{0.1}{25.0} = 0.044 \\ \Rightarrow \frac{Δ g}{g} \times 100 = 4.4 % \end{array}$

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