A 3-kg mass is attached to a spring with stiffness k = 75 N/m, as in Figure 4.1, page 152. The mass is displaced$\frac{\mathbf{1}}{\mathbf{4}}\mathbf{m}$ to the left and given a velocity of 1 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position?

The differential equation is,

$\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

From the given information,

$$\begin{gathered} \mathbf{m}\mathbf{=}\mathbf{3}\mathbf{kg} \\ \mathbf{k}\mathbf{=}\mathbf{75}\frac{\mathbf{N}}{\mathbf{m}} \\ \mathbf{b}\mathbf{=}\mathbf{0} \end{gathered}$$

Substitute the all value of m, k and b in the equation (1),

$\begin{array}{rcl}\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{3}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{0}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{75}\right)\mathbf{y}& \mathbf{=}& \mathbf{0}\\ \mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{75}\mathbf{y}& \mathbf{=}& \mathbf{0} ......\left(\mathbf{2}\right)\end{array}$

The mass is displaced $\frac{1}{4}m$to the left and given a velocity of 1 m/sec to the right.

Therefore,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$

The auxiliary equation for the above equation,

$\begin{array}{rcl}\mathbf{3}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{75}& \mathbf{=}& \mathbf{0}\\ \mathbf{3}{\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{75}\\ {\mathbf{m}}^{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{25}\\ \mathbf{m}& \mathbf{=}& \mathbf{\pm }\mathbf{5}\mathbf{i}\end{array}$

The root of an auxiliary equation is,

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{i}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{i}$

The general solution of the given equation is,

$\mathbf{y}\mathbf{=}\mathbf{Acos}\mathbf{5}\mathbf{t}\mathbf{+}\mathbf{Bsin}\mathbf{5}\mathbf{t} ......\left(\mathbf{3}\right)$

Given the initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$

Substitute the value of $y=-\frac{1}{4}$and $t=0$in the equation (3),

$\begin{array}{rcl}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}& \mathbf{=}& \mathbf{Acos}\mathbf{5}\left(\mathbf{0}\right)\mathbf{+}\mathbf{Bsin}\mathbf{5}\left(\mathbf{0}\right)\\ \mathbf{A}& \mathbf{=}& \mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\end{array}$

Now, find the derivative of the equation (3),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{Asin}\mathbf{5}\mathbf{t}\mathbf{+}\mathbf{5}\mathbf{Bcos}\mathbf{5}\mathbf{t}$

Substitute the value of $y^{\text{'}}=1$and $t=0$in the above equation,

$\begin{array}{rcl}\mathbf{1}& \mathbf{=}& \mathbf{-}\mathbf{5}\mathbf{Asin}\mathbf{5}\left(\mathbf{0}\right)\mathbf{+}\mathbf{5}\mathbf{Bcos}\mathbf{5}\left(\mathbf{0}\right)\\ \mathbf{5}\mathbf{B}& \mathbf{=}& \mathbf{1}\\ \mathbf{B}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{5}}\end{array}$

Substitute the value of A and B in the equation (3),

$\mathbf{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{sin}\mathbf{5}\mathbf{t} ......\left(\mathbf{4}\right)$

Therefore, the amplitude of the motion,

$\begin{array}{rcl}\mathbf{R}& \mathbf{=}& \sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{+}{\mathbf{B}}^{\mathbf{2}}}\\ & \mathbf{=}& \sqrt{{\left(\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{\mathbf{1}}{\mathbf{5}}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{32}\end{array}$

So, the amplitude of the motion is 0.32m

The period of motion is $\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{5}}\mathbf{rad}$and the frequency is $\frac{\mathbf{5}}{\mathbf{2}\mathbf{\pi }}{\mathbf{rad}}^{\mathbf{-}\mathbf{1}}$.

Let after t sec the mass will be at the equilibrium position;

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0}$

From the equation (4),

$\begin{array}{rcl}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{sin}\mathbf{5}\mathbf{t}& \mathbf{=}& \mathbf{0}\\ \mathbf{tan}\mathbf{5}\mathbf{t}& \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{25}\\ \mathbf{t}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{5}}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{1}\mathbf{.}\mathbf{25}\right)\\ \mathbf{t}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{18}\\ & & \end{array}$

Thus, after 0.18 sec the mass will be at the equilibrium position.