Question

The time peroid of oscillation of a simple pendulum in an experiment is observed as 3.60 s , 3.72 s and 3.68 s. Find the percentage error in measured value of time period.

(Ans is 1.2 %)

Answer 1

The measured value of time periods : $t_1=3.60s,t_2=3.72s,t_3=3.68s$

The mean value of time period $t_m=\frac{3.60+3.72+3.68}{3}=3.66sec$

The absolute errors in the measurement are :
$$\begin{gathered} t_m-t_1=3.66-3.60=0.06s \\ {t}_m-t_1=3.66-3.72=-0.06s \\ {t}_m-t_1=3.66-3.68=-0.02s \end{gathered}$$

Mean absolute error
$$\begin{gathered} ∆t_{mean}=\frac{0.06+0.06+0.02}{3}=0.046s \\ Percentageerror=\frac{∆t_{mean}}{t_{mean}}\times 100=\frac{0.046}{3.66}\times 100=1.2percent \end{gathered}$$