Question

A mass weighing $\mathbf{20}\mathbf{\text{pounds}}$stretches a springand another spring $\mathbf{\text{2inches}}$. The two springs are then attached in parallel to a common rigid support in the manner shown in Figure$\mathbf{5}.\mathbf{1}.\mathbf{5}$. Determine the effective spring constant of the double-spring system. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of$\mathbf{\text{2}}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$.

Short answer

Effective spring constant is$160$and equation of motion is $x\left(t\right)=\frac{1}{8}\sin 16t$.

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, $W=20\text{pounds}$

$$\begin{gathered} m=\frac{20}{32}\text{slug}=\frac{5}{8}\text{slug} \\ 6\text{inches}=\frac{1}{2}\text{ft} \\ 2\text{inches}=\frac{1}{6}\text{ft} \end{gathered}$$

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}$

$\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,

$\begin{array}{rcl}k& =& k_1+k_2\\ & =& 40+120\\ & =& 160\end{array}$

Equation of motion:

Differential equation is,

$\begin{array}{rcl}m\frac{d^{2}x}{d{t}^2}& =& -kx\\ \frac{5}{8}\frac{d^{2}x}{d{t}^2}& =& -160x\\ \frac{d^{2}x}{d{t}^2}& =& -256x\\ \frac{d^{2}x}{d{t}^2}+256x& =& 0\end{array}$

The equation of the motion is,

$x\left(t\right)=c_1\cos 16t+c_2\sin 16t$

Apply initial condition:

Mass is initially released from the equilibrium.

$\begin{array}{rcl}x\left(0\right)& =& 0\\ x^{\text{'}}\left(0\right)& =& 2\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\\ c_1\cos 0+c_2\sin 0& =& 0\\ c_1& =& 0\end{array}$

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& -16c_1\sin 16t+16c_2\cos 16t\\ 2& =& x^{\text{'}}\left(0\right)\\ c_2& =& \frac{1}{8}\end{array}$

Hence, equation of the motion is $x\left(t\right)=\frac{1}{8}\sin 16t$.