Question

Solve Problem 13 again, but this time assume that the springs are in series as shown in Figure 5.1.6.

Short answer

Effective spring constant is$30$& equation of motion is $x\left(t\right)=\frac{1}{2\sqrt{3}}\sin 4\sqrt{3}t$.

Step-by-step solution

Definition:

Spring Constant or force constant is defined as the applied force if the displacement in the spring is unity.

Find spring constant:

Weight of mass$W=20\text{pounds}$

$\begin{array}{rcl}m& =& \frac{20}{32}\text{slug}\\ & =& \frac{5}{8}\text{slug}\end{array}$

$6\text{inches}=\frac{1}{2}\text{ft},2\text{inches}=\frac{1}{6}\text{ft}$

Spring constants of two springs are:

$\begin{array}{rcl}{k}_1& =& \frac{W}{s}\\ & =& \frac{20}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ & =& 40\end{array}$

And

$\begin{array}{rcl}{k}_2& =& \frac{20}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}\\ & =& 120\end{array}$

Effective spring constant:

The effective spring constant of the double-spring system is,

$\begin{array}{rcl}k& =& \frac{k_1{k}_2}{k_1+k_2}\\ & =& \frac{\left(40\right)\left(120\right)}{40+120}\\ & =& 30\end{array}$

Equation of motion:

Differential equation is,

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=-kx \\ \frac{5}{8}\frac{d^{2}x}{d{t}^2}=-30x \\ \frac{d^{2}x}{d{t}^2}=-48x \\ \frac{d^{2}x}{d{t}^2}+48x=0 \end{gathered}$$

Equation of the motion is,

$x\left(t\right)=c_1\cos 4\sqrt{3}t+c_2\sin 4\sqrt{3}t$

Apply initial condition:

Mass is initially released from the equilibrium.

$\begin{array}{rcl}\mathrm{x}\left(0\right)& =& 0\\ {\mathrm{x}}^{\text{'}}\left(0\right)& =& 2\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\\ 0& =& {\mathrm{c}}_1\cos 0+{\mathrm{c}}_2\sin 0\\ {\mathrm{c}}_1& =& 0\end{array}$

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& -4\sqrt{3}{c}_1\sin 4\sqrt{3}t+4\sqrt{3}{c}_2\cos 4\sqrt{3}t\\ 2& =& x^{\text{'}}\left(0\right)\\ c_2& =& \frac{1}{2\sqrt{3}}\end{array}$

Hence, the equation of the motion is $x\left(t\right)=\frac{1}{2\sqrt{3}}\sin 4\sqrt{3}t$.