Question

Use the result of Problem 8 to prove that if the pendulum in Figure 4.18 on page 208 is released from rest at the angle $\mathbf{0}\mathbf{<}\mathbf{\alpha }\mathbf{<}\mathbf{\pi }$, then $\left|\mathbf{\theta }\left(\mathbf{t}\right)\right|\le \mathbf{\alpha }$for all t.

Short answer

Therefore, the given statement is true. The pendulum in Figure 4.18 is released from rest at the angle $0<\mathrm{\alpha }<\mathrm{\pi }$, then $\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$for all t is true.

Step-by-step solution

The Energy Integral Lemma

Let $\mathrm{y}\left(\mathrm{t}\right)$be a solution to the differential equation $\mathrm{y}"=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t.

Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$i.e. $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity$\mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant; i.e.role="math" localid="1655365549886" $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$ .

Prove the given equation

Referring from Problem 8: $\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\ell }\mathrm{cos\theta }=\mathrm{constant}\dots \left(1\right)$

To prove: the pendulum is released from rest at the angle $0<\mathrm{\alpha }<\mathrm{\pi }$, then$\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$for all t.

The initial conditions are $\mathrm{\theta }\left(0\right)=\mathrm{\alpha },\mathrm{\theta }\text{'}\left(0\right)=0$.

Let us take role="math" localid="1655365792527" $\mathrm{constant}=\mathrm{k}$. Then,

role="math" localid="1655365833327" $\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\ell }\mathrm{cos\theta }=\mathrm{k}\dots \left(2\right)$

Now, implement the initial conditions here.

role="math" localid="1655365997812" $\begin{array}{c}\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }=\mathrm{k}\\ \frac{1}{2}\mathrm{\theta }\text{'}{\left(0\right)}^2-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }\left(0\right)=\mathrm{k}\\ 0-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }=\mathrm{k}\\ \mathrm{k}=-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }\end{array}$

Now, substitute the value of k in equation (2).

Here ${(\mathrm{\theta }\text{'})}^2$can’t be a negative number. So, we can write this condition as:

$\begin{array}{c}\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }\ge \frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }\\ \mathrm{cos\theta }\ge \mathrm{cos\alpha }\end{array}$

Multiply${\cos }^{-1}$ on both sides.

Hence it is proved that $\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$.