A clock pendulum oscillates at a frequency of 2.5 Hz. At t= 0, it is released from rest starting at an angle of\(12^\circ \)to the vertical. Ignoring friction, what will be the position (angle in radians) of the pendulum at (a)t= 0.25 s, (b) t = 1.60 s, and (c) t = 500 s?

In this problem, firstly determine the relation of angulardisplacement of the pendulum in terms of frequency at any time t and then calculate the position of the pendulum.

The frequency of the simple pendulum is f = 2.5 Hz.

The maximum angular displacement of the simple pendulum at t = 0 is, \({\theta _{\max }} = 12^\circ = \left( {12^\circ } \right)\left( {\frac{{\frac{\pi }{{180}}{\rm{rad}}}}{{1^\circ }}} \right) = \frac{\pi }{{15}}{\rm{rad}}\).

The relation of displacement of a pendulum is given by,

\(x = A\cos \omega t\) ….. (i)

If \(\theta \)is the angle swept by the pendulum of length l at a particular instant of time \(t\), then the arc length can be expressed as the product of angle and the radius\(\left( {x = l\theta } \right)\). So, the equation (i) can be expressed as:

\(\begin{aligned}{c}l\theta &= l{\theta _{\max }}\cos \omega t\\\theta &= {\theta _{\max }}\cos \omega t\end{aligned}\)

If fis the frequency of the simple pendulum, then the angular frequency of the simple pendulum is given as:

\(\omega = 2\pi f\)

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

\(\theta = {\theta _{\max }}\cos \left( {2\pi ft} \right)\)

On substituting the given values in above equation, you will get:

\(\begin{aligned}{l}\theta &= \left( {\frac{\pi }{{15}}{\rm{rad}}} \right)\cos \left( {2\pi \left( {2.5\;{\rm{Hz}}} \right)t} \right)\\\theta &= \left( {\frac{\pi }{{15}}{\rm{rad}}} \right)\cos \left( {5.0\pi t} \right)\end{aligned}\) ….. (ii)

Using (ii), the position of the pendulum at \(t = 0.25\;{\rm{s}}\) is:

\(\begin{aligned}{c}{\theta _{0.25}} &= \left( {\frac{\pi }{{15}}} \right)\cos (5.0\pi (0.25\;{\rm{s}}))\\ &= - 0.148\;{\rm{rad}}\\ &\approx - 0.1{\rm{5}}\;{\rm{rad}}\end{aligned}\)

Therefore, the position of the pendulum at \(t = 0.25\)s is approximately -0.15 rad.

On substituting \(t = 1.60\;{\rm{s}}\) in (ii), you will get:

\(\begin{aligned}{c}{\theta _{1.60}} &= \left( {\frac{\pi }{{15}}} \right)\cos (\left( {5.0\;{\rm{Hz}}} \right)\pi (1.60\;{\rm{s}}))\\ &= \left( {\frac{\pi }{{15}}} \right)\cos 8\pi \\ &= \frac{\pi }{{15}}\;{\rm{rad}}\end{aligned}\)

Therefore, the position of the pendulum at t = 1.60 s is \(\frac{\pi }{{15}}{\rm{rad}}\).

On substituting \(t = 500\;{\rm{s}}\) in (ii), you will get:

\(\begin{aligned}{c}{\theta _{500\;{\rm{s}}}} &= \left( {\frac{\pi }{{15}}} \right)\cos (\left( {5.0\;{\rm{Hz}}} \right)\pi (500\;{\rm{s}}))\\ &= \left( {\frac{\pi }{{15}}} \right)\cos 2500\pi \\ &= \frac{\pi }{{15}}{\rm{rad}}\end{aligned}\)

Therefore, the position of the pendulum at \(t = 500\)s is \(\frac{\pi }{{15}}{\rm{rad}}\).