In an experiment to determine the period of a simple pendulum of length 1 m , it is attached to diff

Question

In an experiment to determine the period of a simple pendulum of length 1 m , it is attached to different spherical bobs of radii $r_{1}$ and $r_{2}$. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5 	imes 10^{-4} \mathrm{~s}$, the difference in radii, $\left|r_{1}-r_{2}ight|$ is best given by\

Professor PC Thomas Classes · Answer

Time period of simple pendulum,\ $T=2 \pi \sqrt{\frac{l}{g}} \quad$ i.e. $T \propto \sqrt{l}$\ $\Rightarrow T=6.28 \sqrt{\frac{1}{9.8}} \simeq 2 \mathrm{~s}$\ $T_{1}-T_{2}=\Delta T=5 	imes 10^{-4} \mathrm{~s}$\ $\left|r_{1}-r_{2}ight|=\Delta l=$ ?\ Use error analysis to find $\left|r_{1}-r_{2}ight|$\ $\frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l} \Rightarrow \Delta l=2 	imes \frac{\Delta T}{T} 	imes l$\ $=2 	imes \frac{5 	imes 10^{-4}}{2} 	imes 1 \mathrm{~m}$\ $=5 	imes 10^{-2} \mathrm{~cm}=0.05 \mathrm{~cm}$\