Chapter 5.4, Problem 45E

a.

To determine

To find the interval.

Answer to Problem 45E

Both masses will cross each other when $t$ is an integral multiple of $\pi$ is $t=k\pi .$

Explanation of Solution

Given position of two masses $s_1=2\sin t,s_2=\sin 2t.$

Calculation: It is given in the question that position of two masses is given by

  $s_1=2\sin t,s_2=\sin 2t.$

Both masses will cross each other when,

  $\begin{array}{l}{s}_1=s_2\\ 2\sin t=\sin 2t\\ 2\sin t=2\sin t\cos t\end{array}$

It occurs for $t=k\pi$ where $k$ is an integer.

Thus, both masses will cross each other when $t$ is an integral multiple of $\pi$ is $t=k\pi$

b.

To determine

To find the vertical distance between masses.

Answer to Problem 45E

The distance is greatest at $t=\frac{4\pi }{3}$ sec and the distance is $\frac{3\sqrt{3}}{2}.$

Explanation of Solution

Given position of two masses $s_1=2\sin t,s_2=\sin 2t.$

Calculation:

Vertical distance between two bodies is,

  $\begin{array}{l}f\left(t\right)=s_1-s_2\\ =\sin 2t-2\sin t\\ =2\sin t\cos t-2\sin t\\ =2\sin t\left(\cos t-1\right)\\ f\left(t\right)=2\sin t\left(\cos t-1\right)\\ f^{\prime }\left(t\right)=2\sin t\left(-\sin t\right)+2\left(\cos t-1\right)\cos t\\ =-2{\sin }^{2}t+2{\cos }^2-2\cos t\\ =2\cos 2t-2\cos t\end{array}$

Critical points occur at

  $\begin{array}{l}{f}^{\prime }\left(x\right)=0\\ 2\cos 2t-2\cos t=0\\ 2\left({\cos }^{2}t-1-\cos t\right)=0\\ 2\left(2\cos t+1\right)\left(\cos t-1\right)=0\\ \Rightarrow \cos t=1,\cos t=\frac{1}{2}\end{array}$

It occurs at $t=\frac{4\pi }{3}$

Here critical point and endpoint lies in $\left(0,0\right),\left(2\pi ,0\right),\left(\frac{2\pi }{3},\frac{3\sqrt{3}}{2}\right),\left(\frac{4\pi }{3},\frac{3\sqrt{3}}{2}\right)$ At $t=\frac{4\pi }{3}.$

  $\begin{array}{l}f\left(t\right)=2\sin t\left(\cos t-1\right)\\ =\frac{3\sqrt{3}}{2}\end{array}$

Thus, the distance is greatest at $t=\frac{4\pi }{3}$ sec and the distance is $\frac{3\sqrt{3}}{2}$ .