The period of oscillation of a pendulum is T = 2 pi

The period of oscillation of a pendulum is $T = 2\pi\sqrt{\dfrac{l}{981}}$ seconds, with $l$ in centimetres. By how much must the length $l = 20\ \text{m}$ be changed so that the period decreases by $0.1\ \text{s}$?

Step 1: Work in the units of the formula

The formula expects centimetres, so $l = 20\ \text{m} = 2000\ \text{cm}$, and

\[ T = 2\pi\sqrt{\frac{2000}{981}} = 2\pi\sqrt{2.038736} = 2\pi\times1.427843 \approx 8.9714\ \text{s}. \]

Step 2: Differentiate

Writing $T = 2\pi\bigl(l/981\bigr)^{1/2}$: \[\frac{dT}{dl} = 2\pi\cdot\frac12\left(\frac{l}{981}\right)^{-1/2}\cdot\frac{1}{981} = \frac{\pi}{981}\sqrt{\frac{981}{l}} = \frac{\pi}{\sqrt{981\,l}}.\]

Step 3: Invert

\[ dl = \frac{\sqrt{981\,l}}{\pi}\,dT. \]

A decrease of the period means $dT = -0.1\ \text{s}$.

Step 4: Evaluate at $l = 2000\ \text{cm}$

\[ 981\times2000 = 1\,962\,000, \qquad \sqrt{1\,962\,000} \approx 1400.714, \]

\[ dl = \frac{1400.714}{\pi}\times\bigl(-0.1\bigr) \approx -44.59\ \text{cm}. \]

Step 5: Verification by relative errors

Since $T\propto\sqrt{l}$, the relative change of the period is half that of the length: \[\frac{dT}{T} = \frac12\,\frac{dl}{l} \;\Longrightarrow\; \frac{dl}{l} = 2\cdot\frac{-0.1}{8.9714} = -0.022293\]

\[dl = -0.022293\times2000 = -44.59\ \text{cm} \;\;\checkmark\]

\[\text{Exact: } T(1955.41) = 2\pi\sqrt{\frac{1955.41}{981}} = 8.8714\ \text{s}, \text{ a fall of exactly } 0.1000\ \text{s} \;\;\checkmark\]