Question

Solve Problem 14 again, but this time assume that the springs are in series as shown in Figure 5.1.6.

Short answer

Weight of first mass$187.5\text{pounds}$.

Step-by-step solution

Hooke’s law:

Hooke's law is a law of physics that states that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance that is,

$\mathit{F}\mathbf{=}\mathit{k}\mathit{x}$

Here, $\mathit{k}$is a constant factor characteristic of the spring, and $\mathit{x}$is small compared to the total possible deformation of the spring.

Find spring constant:

Let the weight of first mass be $W_1$.

First mass stretches one spring $\frac{1}{3}\text{foot}$foot and another spring $\frac{1}{2}\text{foot}$.

Spring constants of two springs are,

$\begin{array}{rcl}{k}_1& =& \frac{W_1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ & =& 3W_1\end{array}$

And

$\begin{array}{rcl}{k}_2& =& \frac{W_1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ & =& 2W_1\end{array}$

Effective spring constant:

Effective spring constants of the double-spring arrangement is,

$\begin{array}{rcl}k& =& \frac{k_1{k}_2}{k_1+k_2}\\ & =& \frac{\left(3W_1\right)\left(2W_1\right)}{3W_1+2W_1}\\ & =& \frac{6}{5}{W}_1\end{array}$

Find angular velocity:

Weight of second mass is $8\text{pounds}$.

$\begin{array}{rcl}m& =& \frac{8}{32}\text{slug}\\ & =& \frac{1}{4}\text{slug}\end{array}$

The differential equation of the motion of the second mass is

$\frac{d^{2}x}{d{t}^2}+w^{2}x=0$

Here, $w^2=\frac{k}{m}$

Time period is $\frac{\pi }{15}\text{sec}$.

Therefore,

$\begin{array}{rcl}\omega & =& \frac{2\pi }{T}\\ & =& \frac{2\pi }{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}\\ & =& 30\end{array}$

Find weight:

Consider the following equation.

${\omega }^2=\frac{k}{m}$

Substitute known values in the above equation.

$$\begin{gathered} {30}^2=\frac{{\displaystyle \frac{6}{5}}{W}_1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ {W}_1=187.5\text{pounds} \end{gathered}$$

Hence, the first mass weighs$187.5\text{pounds}$.