Question 14.7: A simple pendulum consists of a mass hanging on the end of a...
A simple pendulum consists of a mass hanging on the end of a light inextensible string.
The angle θ radians between the string and the vertical satisfies the differential equation
i) Find the length of the pendulum for which the period for small oscillations is 2 s.
The pendulum is released when the string makes an angle of 0.2 rad with the vertical.
ii) Find an equation for θ in terms of t.
iii) Find the times when the pendulum is in the equilibrium position.
iv) Find also the time taken for the pendulum to move from the equilibrium position to a point half-way to the end of the oscillation.
i) The equation of motion, \ddot{θ} = – \frac{g}{l} θ, is the SHM differential equation with ω² = \frac{g}{l}.
All the standard results for SHM apply.
The period is 2 s, so \frac{2π}{ω} = 2 ⇒ ω = π.
Therefore π² = \frac{g}{l}
⇒ l = \frac{g}{π²} = \frac{9.8}{(3.14 …)^{2}}
The length of the pendulum is 0.993 m or about 1 m.
ii) The pendulum is released when θ = 0.2 (θ = 0.2 when t = 0). The most appropriate function to use to model the motion is
θ = 0.2 cos ωt
You have already seen that ω = π.
Therefore θ = 0.2 cos πt
iii) The pendulum is in the equilibrium position when θ = 0, i.e. when
0.2 cos πt = 0
⇒ cos πt = 0
⇒ πt = \frac{π}{2}, \frac{3π}{2}, \frac{5π}{2} ….
⇒ t = \frac{1}{2} , \frac{3}{2}, \frac{5}{2} …
iv) The pendulum is first at the equilibrium position, θ = 0, when t = 0.5.
It is half-way to the end of the oscillation when θ = -0.1 .
The value of t when θ = -0.1 is given by
0.2 cos πt = – 0.1
\begin{matrix} ⇒ πt = \frac{2π}{3} & \boxed{cos \frac{2\pi }{3} = – 0.5} \end{matrix}
⇒ t = \frac{2}{3}.
The time taken from the centre to the half-way point is \frac{2}{3} – \frac{1}{2} = \frac{1}{6} s.
The graph illustrates the first two swings of the pendulum.
