If the length of a simple pendulum is 120cm, calculate its frequency [\(\pi\) = \(\frac{22}{7}\) g = 10ms\(^{-2}\)]

Answer Details

For small oscillations a simple pendulum has period

\[T = 2\pi\sqrt{\frac{L}{g}},\]

and frequency is the reciprocal of period, \(f = 1/T\). Two preparation steps decide whether the arithmetic will be right: the length must be converted to metres, and the frequency must be taken at the end rather than confused with the period.

With \(L = 120\,\mathrm{cm} = 1.20\,\mathrm{m}\), \(g = 10\,\mathrm{m\,s^{-2}}\) and \(\pi = \frac{22}{7}\):

\[\frac{L}{g} = \frac{1.20}{10} = 0.12\,\mathrm{s^{2}}, \qquad \sqrt{0.12} = 0.3464\,\mathrm{s},\] \[T = 2\times\frac{22}{7}\times 0.3464 = 6.286 \times 0.3464 = 2.18\,\mathrm{s}.\]

Hence

\[f = \frac{1}{T} = \frac{1}{2.18} = 0.46\,\mathrm{Hz} \approx 0.5\,\mathrm{Hz}.\]

A useful sense check is that a pendulum about a metre long swings roughly once every two seconds, so its frequency must be about half a hertz. Any answer of a few hertz would mean several complete swings each second, which is physically impossible for a pendulum this long.

Two errors produce the other figures. Leaving the length as \(120\) instead of \(1.20\) inflates \(\sqrt{L/g}\) by a factor of about ten and drives the frequency badly wrong, and stopping at \(T\) and quoting \(2.2\) as though it were the frequency confuses seconds with hertz. Note also that the mass of the bob and the amplitude do not appear in the formula, so they never affect the answer for small swings.