A simple pendulum oscillates with an amplitude of\(10.0^\circ \). What fraction of the time does it spend between\( + 5.0^\circ \)and\( - 5.0^\circ \)? Assume SHM.

If a particle is performing simple harmonic motion (SHM) and starts moving from rest from the point of its maximum displacement (x = A) towards the equilibrium position, then its displacement is given as:

\(x = A\cos \omega t\)

The amplitude of the simple pendulum is \({\theta _0} = 10.0^\circ \).

Consider the displacement of the simple pendulum starts from the position of its maximum displacement in the positive direction. Thus, the angular displacement \(\left( \theta \right)\) at any time t will be represented by the cosine function,

\(\theta = {\theta _0}\cos (\omega t)\)

If Tis the time period of the simple pendulum, then the angular frequency of the simple pendulum is given as:

\(\omega = \frac{{2\pi }}{T}\)

Thus, the angular displacement \(\left( \theta \right)\) at any time t can be written as:

\(\theta = {\theta _0}\cos \left( {\frac{{2\pi }}{T}t} \right)\) ...(i)

If pendulum reaches at \(\theta = 5.0^\circ \) in time t, then using equation (i) the time is calculated as:

\(\begin{array}{5.0^^\circ } = \left( {{{10.0}^^\circ }\cos \left( {\frac{{2\pi t}}{T}} \right)} \right)\\ \frac{{{{5.0}^^\circ }}}{{{{10.0}^^\circ }}} = \left( {\cos \left( {\frac{{2\pi t}}{T}} \right)} \right)\\ \cos \left( {\frac{\pi }{3}} \right) = \left( {\cos \left( {\frac{{2\pi t}}{T}} \right)} \right)\\ t = \left( {\frac{T}{6}} \right)

\end{array}\)

Thus, the time taken by the pendulum to change its position from \(10.0^\circ \) to \(5.0^\circ \)is one-sixth of the time period.

Also, the time taken by the pendulum to change its position from \(10.0^\circ \) to \(0^\circ \)is one-fourth of the time period. Therefore, the time taken to change the position from \(5.0^\circ \) to \(0^\circ \)is:

\(\begin{aligned}t' &= \frac{T}{4} - \frac{T}{6}\\ &= \frac{T}{{12}}\end{aligned}\)

Since the cosine function is symmetric, therefore pendulum takes same time \(t'\) to change its position from \(0^\circ \) to \( - 5.0^\circ \). Thus, in going from \( + 5.0^\circ \) to \( - 5.0^\circ \) , time taken by the pendulum is \(2t'\).

Also, the time taken by the pendulum in coming back from \( - 5.0^\circ \) to \( + 5.0^\circ \) is same as that taken in going from \( + 5.0^\circ \) to \( - 5.0^\circ \). Thus, total time spent by the simple pendulum in between \( + 5.0^\circ \) to \( - 5.0^\circ \) is:

\(\begin{aligned}{c}t'' &= 4t'\\ &= 4\left( {\frac{T}{{12}}} \right)\\ &= \frac{T}{3}\end{aligned}\)

Thus, one-third of the time is spent between \( + 5.0^\circ \) and \( - 5.0^\circ \).