A student uses a simple pendulum of exactly 1 m length to determine $g$, the acceleration due to gra

Question

A student uses a simple pendulum of exactly 1 m length to determine $g$, the acceleration due to gravity. He uses a stopwatch with the least count of 1 s for this and records 40 s for 20 oscillations. For the observation, which of the following statements is correct?\

Professor PC Thomas Classes · Answer

(a): Time period of 1 oscillation, $T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}$ Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$ $ 	herefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T} \quad 	ext { or } \quad \Delta T=\frac{T}{t} \Delta t=\left(\frac{2}{40}ight)(1)=0.05 \mathrm{~s} \ $ Further, $T=2 \pi \sqrt{\frac{l}{g}}$ or $T \propto g^{-1 / 2}$ \ $ 	herefore \quad \frac{\Delta T}{T} 	imes 100=-\frac{1}{2} 	imes \frac{\Delta g}{g} 	imes 100 $ Percentage error in determination of $g$ is $ \ \frac{\Delta g}{g} 	imes 100=-200 	imes \frac{\Delta T}{T}=-\frac{200 	imes 0.05}{2}=-5 \% $