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 $\dfrac{\Delta z}{z}\left( \text{in }\!\!\%\!\!\text{ } \right)$ in $z$ is given by
$\dfrac{\Delta z}{z}\left( \text{in }\!\!\%\!\!\text{ } \right)=\left| m\dfrac{\Delta x}{x}\left( \text{in }\!\!\%\!\!\text{ } \right) \right|+\left| n\dfrac{\Delta y}{y}\left( \text{in }\!\!\%\!\!\text{ } \right) \right|$ --(1)
Where $\dfrac{\Delta x}{x}\left( \text{in }\!\!\%\!\!\text{ } \right),\dfrac{\Delta y}{y}\left( \text{in }\!\!\%\!\!\text{ } \right)$ 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 $\Delta t$ in $t$ is $\Delta t=1s$.
Now, $T=\dfrac{t}{100}$
$\therefore \Delta T=\dfrac{\Delta t}{100}=\dfrac{1}{100}=0.01s$
$\therefore \dfrac{\Delta T}{T}\left( \text{in }\!\!\%\!\!\text{ } \right)=\dfrac{0.01}{0.5}\times 100=0.02\times 100=2\%$ --(2)
Now, the measured length of the pendulum is $l=10cm=0.1m$ $\left( \because 10cm=0.1m \right)$
Now, the error $\Delta l$ in the measured length is the accuracy of the scale, that is,
$\Delta l=1mm=0.001m$ $\left( \because 1mm=0.001m \right)$
$\therefore \dfrac{\Delta l}{l}\left( \text{in }\!\!\%\!\!\text{ } \right)=\dfrac{0.001}{0.1}\times 100=1\%$ --(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
$T=2\pi \sqrt{\dfrac{l}{g}}$
Squaring both sides we get
${{T}^{2}}={{\left( 2\pi \sqrt{\dfrac{l}{g}} \right)}^{2}}=4{{\pi }^{2}}\dfrac{l}{g}$
$\therefore g=4{{\pi }^{2}}\dfrac{l}{{{T}^{2}}}$ --(4)
Using (1) for (4), we get
$\dfrac{\Delta g}{g}\left( \text{in }\!\!\%\!\!\text{ } \right)=\left| \dfrac{\Delta 4{{\pi }^{2}}}{4{{\pi }^{2}}}\left( \text{in }\!\!\%\!\!\text{ } \right) \right|+\left| \dfrac{\Delta l}{l}\left( \text{in }\!\!\%\!\!\text{ } \right) \right|+\left| -2\dfrac{\Delta T}{T}\left( \text{in }\!\!\%\!\!\text{ } \right) \right|$
Putting the values in the above equation we get
$\dfrac{\Delta g}{g}\left( \text{in }\!\!\%\!\!\text{ } \right)=\left| 0 \right|+\left| 1 \right|+\left| -2\left( 2 \right) \right|=0+1+4=5\%$ $\left( \because \Delta 4{{\pi }^{2}}=0,\text{ since it is a constant and does not change} \right)$
Therefore, we have got the required percentage error in the determination of $g$ as $5\%$.

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.