Chapter 5, Problem 127RE

To determine

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

Answer to Problem 127RE

The required model is $y=\frac{\sqrt{10}}{2}\sin \left[8t+{\tan }^{-1}\left(-\frac{1}{3}\right)\right]$ .

Explanation of Solution

Given information:

The model: $y=1.5\sin 8t-0.5\cos 8t$

Formula used:

The sum and difference formula:

  $\sin u\cos v+\cos u\sin v=\sin \left(u+v\right)$

Consider that a weight attached to a spring is suspended vertically from a ceiling.

The given model is,

  $y=1.5\sin 8t-0.5\cos 8t$

Here, y is the distance in feet from equilibrium and t is the time in seconds.

The model can be written as,

  $y=0.5\left(3\sin 8t-\cos 8t\right)$

Multiply and divide the right hand side of the above expression by $\sqrt{3^2+{\left(-1\right)}^2}$ .

  $\begin{array}{c}y=0.5\left(\sqrt{3^2+{\left(-1\right)}^2}\right)\left[\frac{3}{\left(\sqrt{3^2+{\left(-1\right)}^2}\right)}\sin 8t-\frac{1}{\left(\sqrt{3^2+{\left(-1\right)}^2}\right)}\cos 8t\right]\\ y=\frac{1}{2}\left(\sqrt{10}\right)\left(\frac{3}{\sqrt{10}}\right)\sin 8t-\frac{1}{\sqrt{10}}\cos 8t\\ y=\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\sin 8t-\frac{1}{\sqrt{10}}\cos 8t\right)\end{array}$

Assume that $\frac{3}{\sqrt{10}}=\cos \alpha$ and $-\frac{1}{\sqrt{10}}=\sin \alpha$ .

Substitute $\cos \alpha$ for $\frac{3}{\sqrt{10}}$ and $\sin \alpha$ for $-\frac{1}{\sqrt{10}}$ in the above expression.

  $\begin{array}{l}y=\frac{\sqrt{10}}{2}\left(\cos \alpha \sin 8t+\sin \alpha \cos 8t\right)\\ y=\frac{\sqrt{10}}{2}\left(\sin 8t\cos \alpha +\cos 8t\sin \alpha \right)\end{array}$

Use the sum and difference formula $\sin u\cos v+\cos u\sin v=\sin \left(u+v\right)$ on the above expression.

  $y=\frac{\sqrt{10}}{2}\sin \left(8t+\alpha \right)$..............… (1)

Know that $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$ .

Substitute $-\frac{1}{\sqrt{10}}$ for $\sin \alpha$ and $\frac{3}{\sqrt{10}}$ for $\cos \alpha$ in the above expression.

  $\begin{array}{c}\tan \alpha =\frac{-\frac{1}{\cancel{\sqrt{10}}}}{\frac{3}{\cancel{\sqrt{10}}}}\\ \tan \alpha =-\frac{1}{3}\\ \alpha ={\tan }^{-1}\left(-\frac{1}{3}\right)\end{array}$

Now, substitute ${\tan }^{-1}\left(-\frac{1}{3}\right)$ for $\alpha$ in the expression (1).

  $y=\frac{\sqrt{10}}{2}\sin \left[8t+{\tan }^{-1}\left(-\frac{1}{3}\right)\right]$

Hence, the required model is $y=\frac{\sqrt{10}}{2}\sin \left[8t+{\tan }^{-1}\left(-\frac{1}{3}\right)\right]$ .