Question

The period of oscillation of a simple pendulum of length $l$ is given by $T=2\pi \sqrt{l/g}$. The length $l$ is about $10cm$ and is known to $1mm$ accuracy. The period of oscillation is about $0.5s$. The time of $100$ oscillations has been measured with a stop watch of $1s$ resolution. Find the percentage error in the determination of $g$.

Answer 1

Hint: This problem can be solved by writing the mathematical formula for $g$ by using the given formula for $T$ and writing the percentage error in $g$ in terms of the percentage error in $l$ and the percentage error in $T$.

Complete step-by-step answer:
Let us first write the equation for the percentage error in a variable $z$ that is written in terms of two other variables $x$ and $y$. Now, if
$z=x^m{y}^n$
where $m,n$ are real numbers,
The percentage error ${\displaystyle \frac{\mathrm{\Delta }z}{z}}\left(\text{in }\mathrm{\%}\right)$ in $z$ is given by
${\displaystyle \frac{\mathrm{\Delta }z}{z}}\left(\text{in }\mathrm{\%}\right)=\left|m{\displaystyle \frac{\mathrm{\Delta }x}{x}}\left(\text{in }\mathrm{\%}\right)\right|+\left|n{\displaystyle \frac{\mathrm{\Delta }y}{y}}\left(\text{in }\mathrm{\%}\right)\right|$ --(1)
Where ${\displaystyle \frac{\mathrm{\Delta }x}{x}}\left(\text{in }\mathrm{\%}\right),{\displaystyle \frac{\mathrm{\Delta }y}{y}}\left(\text{in }\mathrm{\%}\right)$ are the percentage errors in $x,y$ respectively.
Now, let us analyze the question.
The time period of the pendulum is $T=0.5s$.
Now the time required for $100$ oscillations is $t=T\times 100=0.5\times 100=50s$
Now, since the resolution for the stop watch is $1s$, the error $\mathrm{\Delta }t$ in $t$ is $\mathrm{\Delta }t=1s$.
Now, $T={\displaystyle \frac{t}{100}}$
$\therefore \mathrm{\Delta }T={\displaystyle \frac{\mathrm{\Delta }t}{100}}={\displaystyle \frac{1}{100}}=0.01s$
$\therefore {\displaystyle \frac{\mathrm{\Delta }T}{T}}\left(\text{in }\mathrm{\%}\right)={\displaystyle \frac{0.01}{0.5}}\times 100=0.02\times 100=2\mathrm{\%}$ --(2)
Now, the measured length of the pendulum is $l=10cm=0.1m$ $(\because 10cm=0.1m)$
Now, the error $\mathrm{\Delta }l$ in the measured length is the accuracy of the scale, that is,
$\mathrm{\Delta }l=1mm=0.001m$ $(\because 1mm=0.001m)$
$\therefore {\displaystyle \frac{\mathrm{\Delta }l}{l}}\left(\text{in }\mathrm{\%}\right)={\displaystyle \frac{0.001}{0.1}}\times 100=1\mathrm{\%}$ --(3)
Now, it is given that the time period $T$, length $l$ and acceleration due to gravity $g$ for a simple pendulum are related as
$T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
Squaring both sides we get
$T^2={\left(2\pi \sqrt{{\displaystyle \frac{l}{g}}}\right)}^2=4{\pi }^2{\displaystyle \frac{l}{g}}$
$\therefore g=4{\pi }^2{\displaystyle \frac{l}{T^2}}$ --(4)
Using (1) for (4), we get
${\displaystyle \frac{\mathrm{\Delta }g}{g}}\left(\text{in }\mathrm{\%}\right)=\left|{\displaystyle \frac{\mathrm{\Delta }4{\pi }^2}{4{\pi }^2}}\left(\text{in }\mathrm{\%}\right)\right|+\left|{\displaystyle \frac{\mathrm{\Delta }l}{l}}\left(\text{in }\mathrm{\%}\right)\right|+|-2{\displaystyle \frac{\mathrm{\Delta }T}{T}}\left(\text{in }\mathrm{\%}\right)|$
Putting the values in the above equation we get
${\displaystyle \frac{\mathrm{\Delta }g}{g}}\left(\text{in }\mathrm{\%}\right)=\left|0\right|+\left|1\right|+|-2\left(2\right)|=0+1+4=5\mathrm{\%}$ $(\because \mathrm{\Delta }4{\pi }^2=0,\text{ since it is a constant and does not change})$
Therefore, we have got the required percentage error in the determination of $g$ as $5\mathrm{\%}$.

Note: Students must note that in formula (1), it is imperative that they take the absolute values for the percentage errors of the physical quantities as a function of whom our required physical quantity is written. This is because the error can be positive or negative but we have to write the error in such a way so that all the errors add up and give the maximum relative or percentage errors.