Chapter 2.10, Problem 41E

(a)

To determine

To compare the solution of the pendulum equation for one period from $t=0$ to $t=2\pi$ and to graph the two solutions.

Explanation of Solution

Given:

The given equation is

  $\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

Consider the spring equation-

  $\frac{d^{2}y}{d{t}^2}=-y(t)$

Consider the differential equation of pendulum.

  $\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$

The solution of the pendulum equation for one period from $t=0$ to $t=2\pi$ can be compared as follows-

1. The solution of the pendulum is given by $y(t)=0.1\cos t$
2. The function represented by the pendulum is the sinusoidal function.
3. The function represented by the spring equation is the line represented by y axis.

The graph of the solutions are as follows-

(b)

To determine

To do the same starting from $y(0)=0.2$

Explanation of Solution

Given:

The given equation is

$\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

By equating $\frac{dy}{dt}$ to zero, we get,

  $y(a_2-b_{2}x-c_{2}y)=0$

Thus, we get,

  $a_2-b_{2}x-c_{2}y=0$

By substituting $x=0$ into $a_2-b_{2}x-c_{2}y=0$ , we get,

  $\begin{array}{l}{a}_2-c_{2}y=0\\ \Rightarrow y=\frac{a_2}{c_2}\end{array}$

Now, for $y(0)=0.2$

  $\begin{array}{l}\frac{a_2}{c_2}=0.2\\ a_2=0.2c_2\end{array}$

Thus, the critical point is $(0,0)$ .

(c)

To determine

To do the same starting from $y(0)=0.5$

Explanation of Solution

Given:

The given equation is

$\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

Consider the equation,

By equating $\frac{dy}{dt}$ to zero, we get,

  $y(a_2-b_{2}x-c_{2}y)=0$

Thus, we get,

  $a_2-b_{2}x-c_{2}y=0$

By substituting $x=0$ into $a_2-b_{2}x-c_{2}y=0$ , we get,

  $\begin{array}{l}{a}_2-c_{2}y=0\\ \Rightarrow y=\frac{a_2}{c_2}\end{array}$

Now, for $y(0)=0.2$

  $\begin{array}{l}\frac{a_2}{c_2}=0.2\\ a_2=0.2c_2\end{array}$

Thus, the critical point is $(0,0)$ .

(d)

To determine

To do the same starting from $y(0)=1.0$

Explanation of Solution

Given:

The given equation is

$\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

By equating $\frac{dy}{dt}$ to zero, we get,

  $y(a_2-b_{2}x-c_{2}y)=0$

Thus, we get,

  $a_2-b_{2}x-c_{2}y=0$

By substituting $x=0$ into $a_2-b_{2}x-c_{2}y=0$ , we get,

  $\begin{array}{l}{a}_2-c_{2}y=0\\ \Rightarrow y=\frac{a_2}{c_2}\end{array}$

Now, for $y(0)=1.0$

  $\begin{array}{l}\frac{a_2}{c_2}=1.0\\ a_2=1.0c_2\end{array}$

Thus, the critical point is $(1,1)$ .

(e)

To determine

To do the same starting from $y(0)=1.5$

Explanation of Solution

Given:

The given equation is

$\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

By equating $\frac{dy}{dt}$ to zero, we get,

  $y(a_2-b_{2}x-c_{2}y)=0$

Thus, we get,

  $a_2-b_{2}x-c_{2}y=0$

By substituting $x=0$ into $a_2-b_{2}x-c_{2}y=0$ , we get,

  $\begin{array}{l}{a}_2-c_{2}y=0\\ \Rightarrow y=\frac{a_2}{c_2}\end{array}$

Now, for $y(0)=1.5$

  $\begin{array}{l}\frac{a_2}{c_2}=1.5\\ a_2=1.5c_2\end{array}$

Thus, the critical point is $(0,0)$ .

f.

To determine

To find the time that pendulum takes to swing all the way back in each case and also to tell whether the period of the pendulum depend on the amplitude.

Answer to Problem 41E

  $\text{0}\text{.2π sec}$ , Yes.

Explanation of Solution

Given:

The given equation is

$\frac{d^{2}y}{d{t}^2}=-\sin (y(t))$ And the solution of the spring equation is $y(t)=0.1\cos t$

Calculation:

The time period can be calculated by the following formula as follows-

  $\begin{array}{l}T=2\pi \sqrt{\frac{l}{g}}\\ T=2\pi \sqrt{\frac{0.1}{10}}\\ T=\frac{2\pi }{10}=0.2\pi \end{array}$

Yes, the period depends upon the amplitude, higher the amplitude higher will be the time taken to come back to its initial position.