Chapter 5.4, Problem 79E

a.

To determine

Write the model in the form $y=\sqrt{a^2+b^2}\sin \left(Bt+C\right)$ .

Answer to Problem 79E

  $\boxed{y=\frac{5}{12}\sin \left(2\theta +{\tan }^{-1}\frac{3}{4}\right)}$

Explanation of Solution

Given information:

A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is modeled by

  $y=\frac{1}{3}\sin 2t+\frac{1}{4}\cos 2t$

Where $y$ is the displacement (in feet) from equilibrium of the weight and $t$ is the time (in seconds).

Use the identity $a\sin B\theta +b\cos B\theta =\sqrt{a^2+b^2}\sin \left(B\theta +C\right)$

where $C=\arctan \left(b/a\right),a>0$ , to write the model in the form

  $y=\sqrt{a^2+b^2}\sin \left(Bt+C\right)$ .

Calculation:

Consider the model,

  $y=\frac{1}{3}\sin 2t+\frac{1}{4}\cos 2t$ ......... $\left(1\right)$

From the identity, $a\sin B\theta +b\cos B\theta =\sqrt{a^2+b^2}\sin \left(B\theta +C\right)$ .......... $\left(2\right)$

Compare $\left(1\right)$ and $\left(2\right)$ then $a=\frac{1}{3},B=2,b=\frac{1}{4}$

  $\begin{array}{l}\sqrt{a^2+b^2}=\sqrt{\frac{1}{9}+\frac{1}{16}}\\ =\sqrt{\frac{25}{9\times 16}}\\ =\frac{5}{12}\end{array}$

And ,

  $\begin{array}{l}C={\tan }^{-1}\left(\frac{b}{a}\right)where,a>0\\ C={\tan }^{-1}\left(\frac{1}{4}\times \frac{3}{1}\right)\\ ={\tan }^{-1}\left(\frac{3}{4}\right)\end{array}$

Thus $y=\frac{1}{3}\sin 2t+\frac{1}{4}\cos 2t$ can be written as ,

  $\begin{array}{l}y=\sqrt{a^2+b^2}\sin \left(B\theta +C\right)\\ y=\frac{5}{12}\sin \left(2\theta +{\tan }^{-1}\frac{3}{4}\right)\end{array}$

Hence, the given model can be written as $\boxed{y=\frac{5}{12}\sin \left(2\theta +{\tan }^{-1}\frac{3}{4}\right)}$ .

b.

To determine

Find the amplitude of the oscillations of the weight.

Answer to Problem 79E

  $\boxed{Amplitude=\frac{5}{12}ft}$

Explanation of Solution

Given information:

A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is modeled by

  $y=\frac{1}{3}\sin 2t+\frac{1}{4}\cos 2t$

Where $y$ is the displacement (in feet) from equilibrium of the weight and $t$ is the time (in seconds).

Find the amplitude of the oscillations of the weight.

Calculation:

Consider the model,in the form of ,

  $y=a\sin \left(\omega t+c\right)$ compare with $y=\frac{5}{12}\sin \left(2\theta +{\tan }^{-1}\frac{3}{4}\right)$

  $Amplitude=\frac{5}{12}$

Hence , $\boxed{Amplitude=\frac{5}{12}ft}$

c.

To determine

Find the frequency of the oscillations of the weight.

Answer to Problem 79E

  $\boxed{\frac{1}{\pi }cycle/\sec }$

Explanation of Solution

Given information:

A weight is attached to a spring suspended vertically from a ceiling. When a driving force is applied to the system, the weight moves vertically from its equilibrium position, and this motion is modeled by

  $y=\frac{1}{3}\sin 2t+\frac{1}{4}\cos 2t$

Where $y$ is the displacement (in feet) from equilibrium of the weight and $t$ is the time (in seconds).

Find the frequency of the oscillations of the weight.

Calculation:

Consider the model for period,

  $y=a\sin \left(\omega t+c\right)$ is $\frac{2\pi }{\omega }$

Period is $\frac{2\pi }{\omega }=\pi$

Frequency is $\frac{1}{period}$

Hence, frequency is $\boxed{\frac{1}{\pi }cycle/\sec }$ .