Convert unit measures for a problem from zill book

Question

I have some questions about a solution for a problem from Zill book "A first course in differential equations 10th edition". I attached an image with the problem PROBLEM AND SOLUTION.

I have the solution, but I don't understand it, I don't understand the part which implies the conversion of unit measures. I know that I can take $g = 32\cdot\frac{ft}{s^2}$, but then what? So my questions are:

1. how to compute (convert unit measures) $\frac{4 pound}{32 \cdot \frac{ft}{s^2}}$ to become $\frac{1}{8}$ ? pound is a unit measure for weight, ft is for distance (like meter), s is for time. So, how they (in image) simplify that raport - division?
2. Then, for the next step: I know that $\omega^2 = \frac{k}{m}$, with $m =\frac{1}{8}slug~~$ and $k = 16\frac{lb}{ft}$. But what about this division: $$\frac{k}{m} = \frac{16\frac{lb}{ft}}{\frac{1}{8}slug} = 128 \frac{lb \cdot slug}{ft} $$ And again, how can I make this division? lb and slug are units of measure for weight and ft is for distance. How can I reduce them?

Answer 1

As I said in my comment, lb is a measure for weight, i.e., a measure for force, while slug is a measure for mass. The problem is that in common talk we do not make a distinction between them. Given that

$$1 \;\mathit{slug}=1 \;\mathit{lb}\;\frac{s^2}{\mathit{ft}},$$

it's easy to obtain

$$\frac{16\frac{lb}{ft}}{\frac{1}{8}slug}=128\frac{\frac{lb}{ft}}{\mathit{lb}\;\frac{s^2}{\mathit{ft}}}=128\frac{1}{s^2}.$$

As I'm not a physicist myself, I don't know if these are the units to be expected.

Answer 2

Were the problem and solution written by different people? There's no need to commit to a new mass unit if you know the relations between weight ($F_g$), mass ($m$), gravitational acceleration ($g$), angular frequency ($\omega$), the spring constant ($k$), and the period ($T$):

$$\begin{align} F_g &= mg \\ \omega^2 &= \frac{k}{m} \\ T &=\frac{2\pi}{\omega} \text{.} \end{align}$$

That's because you can eliminate $m$ and $\omega$ to get

$$\begin{split} T &= 2\pi \sqrt{\frac{F_g}{gk}} \\ &\approx 2\pi \sqrt{\frac{ 4\,\mathrm{lb_f}}{ (32\,\mathrm{ft}\,\mathrm{s}^{-2})(16 \,\mathrm{lb_f}\,\mathrm{ft}^{-1})}} \\ &\approx \boxed{0.6 \,\mathrm{s}}\text{.} \end{split}$$

The solution you provided has carelessly lost units.