Question

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

• A. Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds.
• B. Error $\Delta T$ in measuring $T$, the time period, is 1 second.
• C. Percentage error in the determination of $g$ is 5%.
• D. Percentage error in the determination of $g$ is 2.5%.

Answer 1

Answer: (A), (C)

Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds.
Percentage error in the determination of $g$ is 5%.

We know that, $\frac{\Delta T}{T}=\frac{\Delta t}{t}=\frac{1}{40}$
$T=2$sec/ Oscillation
$\Delta T=0.05\sec$

We have $g=\frac{4{\pi }^{2}L{n}^2}{T^2}$
$\frac{\Delta g}{g}=\frac{2\Delta T}{T}$
$\Rightarrow$ % error $=\frac{2\Delta T}{T}\times 100$

We know, $\frac{\Delta T}{T}=\frac{1}{40}$
$\frac{2\Delta T}{T}\times 100=5\%$