Chapter 2, Problem 32SP

(a)

To determine

To find: The derivative $x(t).$

Answer to Problem 32SP

  $v(t)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

Differentiating $x(t).$ with respect to $t$ ,

  $\begin{array}{l}x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)\\ \frac{d}{dt}x(t)=\frac{d}{dt}\left\{2.0\cos \left(\frac{2\pi t}{3.2}\right)\right\}\\ =-2\sin \left(\frac{2\pi t}{3.2}\right)\left(\frac{2\pi }{3.2}\right)\\ v(t)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)---(2)\end{array}$

The rate of change of position of object represents velocity $v(t)$ .

(b)

To determine

To find: The second derivative $x(t).$

Answer to Problem 32SP

  $a(t)=-\frac{4\pi }{3.2}\left(\frac{2\pi }{3.2}\right)\cos \left(\frac{2\pi t}{3.2}\right)$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

Differentiating $x(t).$ with respect to $t$ ,

  $\begin{array}{l}x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)\\ \frac{d}{dt}x\text{'}\text{'}(t)=\frac{d}{dt}v\text{'}(t)\\ =\frac{d}{dt}\left\{-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)\right\}\\ a(t)=-\frac{4\pi }{3.2}\left(\frac{2\pi }{3.2}\right)\cos \left(\frac{2\pi t}{3.2}\right)---(3)\end{array}$

The rate of change of velocity of object represents velocity $a(t)$ .

(c)

To determine

To find: The position and velocity at $t=0$

Answer to Problem 32SP

  $x(0)=2;v(0)=0$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

  $v(t)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)---(2)$

Substitute $t=0$ in equation (1) and (2),

  $x(0)=2.0\cos \left(\frac{2\pi 0}{3.2}\right)=2$

  $v(0)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi 0}{3.2}\right)=0$

Hence, $x(0)=2;v(0)=0$

(d)

To determine

To find: The position and velocity at $t=1.6$

Answer to Problem 32SP

  $x(1.6)=-2;v(1.6)=0$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

  $v(t)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)---(2)$

Substitute $t=1.6$ in equation (1) and (2),

  $x(1.6)=2.0\cos \left(\frac{2\pi (1.6)}{3.2}\right)=2.0\cos (\pi )=-2$

  $v(1.6)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi (1.6)}{3.2}\right)=-\frac{4\pi }{3.2}\sin (\pi )=0$

Hence, $x(1.6)=-2;v(1.6)=0$

(e)

To determine

To find: The position and velocity at $t=3.2$

Answer to Problem 32SP

  $x(3.2)=2;v(3.2)=0$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

  $v(t)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi t}{3.2}\right)---(2)$

Substitute $t=3.2$ in equation (1) and (2),

  $x(3.2)=2.0\cos \left(\frac{2\pi (3.2)}{3.2}\right)=2.0\cos (2\pi )=2$

  $v(3.2)=-\frac{4\pi }{3.2}\sin \left(\frac{2\pi (3.2)}{3.2}\right)=-\frac{4\pi }{3.2}\sin (2\pi )=0$

Hence, $x(3.2)=2;v(3.2)=0$

(f)

To determine

To find: The differential equation does the object follows.

Answer to Problem 32SP

  $\frac{d^{2}x(t)}{d{t}^2}=-3.855x(t)$

Explanation of Solution

Given information:

Consider the equation, $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$

Calculation:

The position of the object attached to the spring is,

  $x(t)=2.0\cos \left(\frac{2\pi t}{3.2}\right)$ ----- (1)

  $a(t)=-\frac{4\pi }{3.2}\left(\frac{2\pi }{3.2}\right)\cos \left(\frac{2\pi t}{3.2}\right)$ ---- (3)

  $\begin{array}{l}\frac{d^2}{d{t}^2}x(t)=-\frac{4\pi }{3.2}\left(\frac{2\pi }{3.2}\right)\frac{x(t)}{2}from(1)\\ \frac{d^{2}x(t)}{d{t}^2}=-\frac{4\pi }{3.2}\left(\frac{\pi }{3.2}\right)x(t)\end{array}$

Thus the differential equation is $\frac{d^{2}x(t)}{d{t}^2}=-3.855x(t)$