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The time period $T$ of a pendulum of length $l$ is given by $T=2\pi\sqrt{\frac{l}{g}}$. What is the percentage error in $T$ when the error in length is $1\text{%}$.

Attempt-

$\displaystyle T=2\pi\sqrt{\frac{l}{g}}$ and $\displaystyle\frac{\Delta l}{l}=0.01$

By using differentials we get

$\displaystyle\Delta T\approx\frac{dT}{dl}\Delta l=\frac{\pi}{\sqrt{gl}}0.01$

Therefore $\displaystyle\frac{\Delta T}{T}*100=\frac{1}{2l}$ which is wrong dimensionally by an extra $l$ in denominator.

What is the problem here?

Lalit Tolani
  • 3,426

1 Answers1

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${\Delta l} = 0.01*l$ . You have missed out an extra $l$

Therefore Percentage error in $\displaystyle T=\frac{dT}{dl}\Delta l=\frac{dT}{dl}*0.01*l$

Now the $l$ in denominator cancels

To avoid such mistakes you can use logarithmic differentiation also.