Physics - SHM Question with Solution | TestHub

The period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$.

Squaring both sides and rearranging for $g$, we get:

$g=4{\pi }^2\frac{l}{T^2}$

The percentage error in $g$ is given by:

$\frac{\Delta g}{g}\times 100=\left(\frac{\Delta l}{l}+2\frac{\Delta T}{T}\right)\times 100$

Given $l=100\text{ cm}$ and $\Delta l=1\text{ mm}=0.1\text{ cm}$, the relative error in $l$ is:

$\frac{\Delta l}{l}=\frac{0.1}{100}=0.001$

The total time for 100 oscillations is $t=100\times 2\text{ s}=200\text{ s}$.

The least count of the stopwatch is $\Delta t=0.1\text{ s}$.

The relative error in the time period $T$ is equal to the relative error in the total time $t$:

$\frac{\Delta T}{T}=\frac{\Delta t}{t}=\frac{0.1}{200}=0.0005$

Substituting these values into the percentage error formula:

$\frac{\Delta g}{g}\times 100=\left(0.001+2\times 0.0005\right)\times 100$

$\frac{\Delta g}{g}\times 100=(0.001+0.001)\times 100=0.002\times 100=0.2\%$

Answer: $0.2\%$