Chapter 4.4, Problem 45E

(a)

To determine

To find: The time at which the masses $m_1$ and $m_2$ pass each other.

Answer to Problem 45E

The two masses $m_1$ and $m_2$ pass each other at $t=k\pi$ , where k is any integer.

Explanation of Solution

Given information: The position of the masses $m_1$ and $m_2$ are given as $s_1=2\sin t$ and $s_2=\sin 2t$ respectively.

Concept used: Masses $m_1$ and $m_2$ will pass each other when the distances $s_1$ and $s_2$ for the corresponding masses will be equal.

Calculation:

Equate both the given distances, that is, $s_1=s_2$ and solve for time $t$ .

  $\begin{array}{c}2\sin t=\sin 2t\\ 2\sin t=2\sin t\cos t\\ 2\sin t-2\sin t\cos t=0\\ 2\sin t\left(1-\cos t\right)=0\\ \sin \left(t\right)=0\text{ or }\cos \left(t\right)-1=0\\ t=k\pi \end{array}$

  $t=k\pi$ is the common solution to the obtained factors, where $k$ is any integer.

Therefore, both the masses $m_1$ and $m_2$ will pass each other when $t$ is an integral multiple of $\pi$ .

(b)

To determine

To find: The time at which the vertical distance between the masses is the greatest in the interval $0\le t\le 2\pi$ .

Answer to Problem 45E

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

Explanation of Solution

Given information: The position of the masses $m_1$ and $m_2$ are given as $s_1=2\sin t$ and $s_2=\sin 2t$ respectively.

Concept used:

The maximum value of the any function will be at the critical point of that function.

Calculation:

Determine the vertical distance between masses $m_1$ and $m_2$ by subtracting $s_1$ from $s_2$ .

  $\begin{array}{c}d\left(t\right)=s_2-s_1\\ =\sin 2t-2\sin t\\ =2\sin t\cos t-2\sin t\\ =2\sin t\left(\cos t-1\right)\end{array}$

Differentiate the obtained function on both sides with respect to t.

  $\begin{array}{c}\frac{d}{dt}\left(d\left(t\right)\right)=\frac{d}{dt}\left(2\sin t\left(\cos t-1\right)\right)\\ d^{\prime }\left(t\right)=\left(\cos t-1\right)\frac{d}{dt}\left(2\sin t\right)+2\sin t\frac{d}{dt}\left(\cos t-1\right)\\ =\left(\cos t-1\right)2\cos t+2\sin t\left(-\sin t-0\right)\\ =2{\cos }^{2}t-2\cos t-2{\sin }^{2}t\\ =2\left({\cos }^{2}t-{\sin }^{2}t\right)-2\cos t\\ =2\cos 2t-2\cos t\end{array}$

Equate the obtained derivative to zero and solve for t to get the critical point.

  $\begin{array}{c}2\cos 2t-2\cos t=0\\ 2\left(2{\cos }^{2}t-1-\cos t\right)=0\\ 2{\cos }^{2}t-\cos t-1=0\\ 2{\cos }^{2}t-2\cos t+\cos t-1=0\\ 2\cos t\left(\cos t-1\right)+1\left(\cos t-1\right)=0\\ \left(\cos t-1\right)\left(2\cos t+1\right)=0\\ \cos t=1,\cos t=-\frac{1}{2}\end{array}$

The only possible solution in the interval $0\le t\le 2\pi$ occurs at $t=\frac{4\pi }{3}$ . As a result, the critical point in the given interval is at $t=\frac{4\pi }{3}\text{sec}$ .

Differentiate the obtained first derivative on both sides with respect to t.

  $\begin{array}{c}\frac{d}{dt}\left(d^{\prime }\left(t\right)\right)=\frac{d}{dt}\left(2\cos 2t-2\cos t\right)\\ d^{″}\left(t\right)=2\frac{d}{dt}\left(\cos 2t\right)-2\frac{d}{dt}\left(\cos t\right)\\ =-4\sin 2t+2\sin t\end{array}$

Determine the value of the obtained second derivative at $t=\frac{4\pi }{3}$ .

  $\begin{array}{c}{d}^{″}\left(\frac{4\pi }{3}\right)=-4\sin 2\left(\frac{4\pi }{3}\right)+2\sin \left(\frac{4\pi }{3}\right)\\ =-4sin\left(\frac{8\pi }{3}\right)+2sin\left(\frac{4\pi }{3}\right)\\ =-4\cdot \frac{\sqrt{3}}{2}+2\left(-\frac{\sqrt{3}}{2}\right)\\ =-3\sqrt{3}\\ <0\end{array}$

Since the value of the obtained second derivative at $t=\frac{4\pi }{3}$ is negative or less than zero, the vertical distance will be maximum or greatest at $t=\frac{4\pi }{3}\text{sec}$ .

Determine the value of the distance at $t=\frac{4\pi }{3}$ by substituting $t=\frac{4\pi }{3}$ into $d\left(t\right)=2\sin t\left(\cos t-1\right)$ .

  $\begin{array}{l}d\left(\frac{4\pi }{3}\right)=2\sin \left(\frac{4\pi }{3}\right)\left(\cos \left(\frac{4\pi }{3}\right)-1\right)\\ =2\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}-1\right)\\ =-\sqrt{3}\left(-\frac{3}{2}\right)\\ =\frac{3\sqrt{3}}{2}\text{m}\end{array}$

Therefore, the required distance is $\frac{3\sqrt{3}}{2}\text{m}$