Question

In an experiment of simple pendulum, the time period measured was $50$s for $25$ vibrations when the length of the simple pendulum was taken $100$cm. If the least count of stop watch is $0.1$s and that of meter scale is $0.1$cm. Calculate the maximum possible error in the measurement of value of $g$. If the actual value of $g$ at the place of experiment is $9.7720m{s}^{-2}$, Calculate the percentage error.

Answer 1

Sol.The period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ or $T^2=\frac{4{\pi }^{2}l}{g}$ or $g=\frac{4{\pi }^{2}l}{T^2}$As $4$ and $\pi$ are constant, the maximum permissible error in $g$ is given by $\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$here $\Delta L=0.1cm,L=1m=100cm,$ $\Delta T=0.1s,T=50s.$$\therefore \frac{\Delta g}{g}=\frac{0.1}{100}+2\left(\frac{0.1}{50}\right)=\frac{0.1}{100}+\left(\frac{0.1}{25}\right);$$\frac{\Delta g}{g}\times =[\frac{0.1}{100}+\frac{0.1}{25}]\times 100=0.1+0.4=0.5\%.$Now $g=\frac{4{\pi }^2{l}^2}{T^2}.$ Here $T=\frac{50}{25}=2.$ Therefore,$g^{\prime }=\frac{4\times (3.14{)}^4\times (1{)}^2}{(2{)}^2}=9.8596m{s}^{-2}$;Actual value $g=9.7720m{s}^{-1}$. ThereforePercentage error$=\frac{g^{\prime }-g}{g}\times 100$$=\frac{9.8596-9.7720}{9.7720}\times 100=0.8964\%$