Consider a pendulum that is released from rest from an initial displacement of
\(\theta_{0}\) radians. Solving the linear model (7) subject to the initial
conditions \(\theta(0)=\theta_{0}, \theta^{\prime}(0)=0\) gives
\(\theta(t)=\theta_{0} \cos \sqrt{g} \| t .\) Theperiod of oscillations
predicted by this modelisgivenbythefamiliarformula \(T=2 \pi / \sqrt{g l l}=2
\pi \sqrt{U g}\). The interesting thing about this formula for \(T\) is that it
does not depend on the magnitude of the initial displacement \(\theta_{0}\). In
other words, the linear model predicts that the time that it would take the
pendulum to swing from an initial displacement of, say, \(\theta_{0}=\pi /
2\left(=90^{\circ}\right)\) to \(-\pi / 2\) and back again would be exactly the
same time to cycle from, say, \(\theta_{0}=\pi / 360\left(=0.5^{\circ}\right)\)
to \(-\pi / 360\). This is intuitively unreasonable; the actual period must
depend on \(\theta_{0}\).
If we assume that \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) and \(l=32 \mathrm{ft}\),
then the period of oscillation of the linear model is \(T=2 \pi \mathrm{s}\).
Let us compare this last number with the period predicted by the nonlinear
model when \(\theta_{0}=\pi / 4\). Using a numerical solver that is capable of
generating hard data, approximate the solution of
$$\frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, \quad \theta(0)=\frac{\pi}{4},
\quad \theta^{\prime}(0)=0$$
for \(0 \leq t \leq 2\). As in Problem 24 , if \(t_{1}\) denotes the first time
the pendulum reaches the position \(O P\) in Figure 3.11.3, then the period of
the nonlinear pendulum is \(4 t_{1} .\) Here is another way of solving the
equation \(\theta(t)=0\). Expeniment with small step sizes and advance the time
staning at \(t=0\) and ending at \(t=2\). From your hard data, observe the time
\(t_{1}\) when \(\theta(t)\) changes, for the first time, from positive to
negative. Use the value \(t_{1}\) to determine the true value of the period of
the nonlinear pendulum. Compute the percentage relative error in the period
estimated by \(T=2 \pi\).
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