Physics - SHM Question with Solution | TestHub

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

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

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

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 that $l=100$ cm and $\Delta l=1$ mm $=0.1$ cm, the percentage error in length is:

$\frac{\Delta l}{l}\times 100=\frac{0.1}{100}\times 100=0.1\%$

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

The error in measuring the total time is the least count of the stopwatch, so $\Delta t=0.1$ s.

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

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

The percentage error contributed by the time period measurement is:

$2\frac{\Delta T}{T}\times 100=2\times \frac{0.1}{200}\times 100=0.1\%$

Adding the percentage errors, the total percentage error in $g$ is:

$\frac{\Delta g}{g}\times 100=0.1\%+0.1\%=0.2\%$

Answer: $0.2\%$