Question

Consider a pendulum that is released from rest from an initial displacement of radians. Solving the linear model (7) subject to the initial condition $\theta \left(0\right)={\theta }_0,{\theta }^{\text{'}}\left(0\right)=0$gives $\theta \left(t\right)={\theta }_0\cos \sqrt{\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$lt$}\right.}$ . The period of oscillations predicted by this model is given by the familiar formula $T=\raisebox{1ex}{$2\pi $}\!\left/ \!\raisebox{-1ex}{$\sqrt{{\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$I$}\right.}}$}\right.=2\pi \sqrt{\raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$g$}\right.}$.

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 it would take the pendulum to swing from an initial displacement of, say,${\theta }_0=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\left(={90}^o\right)to\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and back again would be exactly the same as the time it would take to cycle from, say,${\theta }_0=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$360\left(=0.5^o\right)to{\displaystyle \raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{${360}^o$}\right.}$}\right.$. This is intuitively unreasonable; the actual period must depend on ${\theta }_0$.

If we assume that $g=32ft/s^2$and $l=32ft$, then the period of oscillation of the linear model is $T=2\pi s$. Let us compare this last number with the period predicted by the non linear model when ${\theta }_0=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$. Using a numerical solver that is capable of generating hard data, approximate the solution of,

$\frac{d^2\theta }{d{t}^2}+\sin \theta =0,\theta \left(0\right)=\frac{\pi }{4},{\theta }^{\text{'}}\left(0\right)=0$

On the interval $0⩽t⩽2$. As in the problem 25, if$t_1$ denotes the first time the pendulum reaches the position OP in Figure 5.3.3, then the period of the non linear pendulum is $4t_1$. Here is another way of solving the equation$\theta \left(t\right)=0$ . Experiment with small step sizes and advance the time, starting at $t=0$and ending at$t=2$ . From your hard data observe the time$t_1$ when$\theta \left(t\right)$ changes , for the first time , from positive to negative. Use the value $t_1$to determine the true value of the period of the non linear pendulum. Compute the percentage relative error in the period estimated by $T=2\pi$.

Short answer

The period is approximately$6.536$. The percentage relative error is$0.00253\%$.

Step-by-step solution

Table obtained as result of running the code in GNU Octave.

To complete this task, we will use GNU Octave. In the next few tables, there will be given time in the left column and the approximate values of $\theta \left(t\right)$at the right column

where $\theta \left(t\right)$is the solution of the given initial value problem

$\frac{d^2\theta }{d{t}^2}+\sin \theta =0,\theta \left(0\right)=\frac{\pi }{4},{\theta }^{\text{'}}\left(0\right)=0$

The codes are given below as,

Finding the percentage relative error

Check in the table when$\theta \left(t\right)$, changes its sign. From the last one, we can see that this will approximately be at$t_1\approx 1.634$. Therefore, the period is approximately$4t_1\approx 6.536$.

Let us compute the percentage relative error in the period estimated by $T=2\pi$.

$\frac{6.536-6.283}{100}=0.00253\%$

Codes used in the GNU Octave

$$\begin{gathered} \backslash \\ functionf=f_{-}26\left(t,y\right) \\ f\left(1\right)=y\left(2\right); \\ f\left(2\right)=-\sin \left(y\left(1\right)\right); \\ endfunction \\ function\left[x,y\right]=od{j}_{-}rk4v\left(f,a,b,y\theta ,n\right) \\ x=linspace\left(a,b,n+1\right); \\ h=x\left(2\right)-x\left(1\right); \\ y=zeros\left(n+1,length\left(y\theta \right)\right); \end{gathered}$$

$$\begin{gathered} y\left(1,:\right)=y\theta ; \\ fori=1:n \\ k1=f\left(\left(x\left(i\right),y\left(i,:\right)\right)\right); \\ k2=f\left(\left(x\left(i\right)+h/2,y\left(i,:\right)+h/2*k1\right)\right); \\ k3=f\left(\left(x\left(i\right)+h/2,y\left(i,:\right)+h/2*k2\right)\right); \\ k4=f\left(\left(x\left(i+1\right),y\left(i,:\right)+h*k3\right)\right); \\ y\left(i+1,:\right)=y\left(i,:\right)+h/6*\left(k1+k2+k3+k4\right) \\ end \\ end \\ \backslash \\ clear; \\ y\theta =\left[\raisebox{1ex}{$pi$}\!\left/ \!\raisebox{-1ex}{$4$}\right.;\theta \right] \\ n=2000 \\ t\theta =\theta \\ T=2 \\ \left[t,y\right]=odi\_rk4v\left(@f\_26,t\theta ,T,y\theta ,n\right) \\ fori=1:n \\ printf\left(\text{'}\%f,\%f\backslash n\text{'},t\left(i\right),\left(i,1\right)\right) \\ endfor \end{gathered}$$

The period is approximately$6.536$. The percentage relative error is $0.00253\%$.