Question

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
(a)1%
(b) 2%
(c) 3%
(d) 4%

Answer 1

(a) 1%
Let l be the length if the pendulum and T be the period.

$$\begin{gathered} \mathrm{Also},\mathrm{let}∆l\mathrm{be}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{length}\mathrm{and}∆T\mathrm{be}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{period}. \\ \mathrm{We}\mathrm{have} \\ \frac{∆l}{l}\times 100=2 \\ \Rightarrow \frac{dl}{l}\times 100=2 \\ \mathrm{Now},T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log T=\log 2\mathrm{\pi }+\frac{1}{2}\log l-\frac{1}{2}\log \mathrm{g} \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{1}{T}\frac{dT}{dl}=\frac{1}{2l} \\ \Rightarrow \frac{dT}{dl}=\frac{T}{2l} \\ \Rightarrow \frac{dl}{l}\times 100=2\frac{dT}{T}\times 100 \\ \Rightarrow \frac{dT}{T}\times 100=\frac{2}{2} \\ \Rightarrow \frac{∆T}{T}\times 100=1 \\ \mathrm{Hence},\mathrm{there}\mathrm{is}\mathrm{an}\mathrm{error}\mathrm{of}1\%\mathrm{in}\mathrm{calculating}\mathrm{the}\mathrm{period}\mathrm{of}\mathrm{the}\mathrm{pendulum}. \end{gathered}$$