Question

Determine the equation of motion if the mass in Problem 3 is initially released from the equilibrium position with a downward velocity of $\mathbf{2}\mathbf{ft}\mathbf{/}\mathbf{s}$.

Short answer

So, the equation is $x\left(t\right)=\frac{\sqrt{6}}{12}\sin 4\sqrt{6}t$.

Step-by-step solution

Definition of Hooke’s law

Hooke's law is a law of physics that states that the force $\mathbf{\left(}\mathit{F}\mathbf{\right)}$needed to extend or compress a spring by some distance $\mathbf{\left(}\mathit{x}\mathbf{\right)}$scales linearly with respect to that distance—that is, ${\mathit{F}}_{\mathbf{s}}\mathbf{=}\mathit{k}\mathit{x}$, where role="math" localid="1667910224364" $\mathit{k}$is a constant factor characteristic of the spring (i.e., its stiffness), and $\mathit{x}$ is small compared to the total possible deformation of the spring.

From the problem (3) determine the equation of motion.

The mass of weighing 24 pounds, the weight stretches the spring 4 inches.

By Hooke's law,$W=kl$

$24=k\left(\frac{1}{3}\right)$Since 4 inches $=\frac{1}{3}$feet

$k=72$

The mass of the object is,$m=\frac{W}{g}$

$m=\frac{24}{32}$Here,$g=32\text{ft}/{\text{sec}}^2$

$=\frac{3}{4}$

There is no damping force applied so$b=0$ .

Differentiation of the equation

The differential equation for this system is,$m\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+kx=0$

Substitute$m=\frac{3}{4},b=0$and$k=72$in the above equation.

$\frac{3}{4}\frac{d^{2}y}{d{x}^2}+0\frac{dy}{dx}+72x=0$

$\frac{d^{2}y}{d{x}^2}+96x=0$

Write the initial conditions:

The mass is released from the equilibrium position with a downward velocity of $2\text{ft}/\text{sec}$.

Thus,$x\left(0\right)=0$

$x^{\text{'}}\left(0\right)=2$

Solution of differential equation

The mass is released from the equilibrium position with a downward velocity of$2ft/\text{sec}$ .

Thus,$x\left(0\right)=0$

$x^{\text{'}}\left(0\right)=2$

The equation of differential equation is,$\frac{d^{2}x}{d{t}^2}+96x=0$ with $x\left(0\right)=0$and $x^{\text{'}}\left(0\right)=2$.

The auxiliary equation is,$m^2+96=0$

$m^2=-96$

$m=\pm i4\sqrt{6}$

The solution of differential equation is,$x\left(t\right)=c_1\cos 4\sqrt{6}t+c_2\sin 4\sqrt{6}t$

Differentiate the$x\left(t\right)$ with respect to $t$.

$x^{\text{'}}\left(t\right)=4\sqrt{6}[-c_1\sin 4\sqrt{6}t+c_2\cos 4\sqrt{6}t]$

Substitute the initial conditions

Putting$x\left(0\right)=0$ and$x^{\text{'}}\left(0\right)=2$

$x\left(0\right)=c_1\cos 0+c_2\sin 0$

$0=c_1$

$x^{\text{'}}\left(0\right)=4\sqrt{6}(-c_1\sin 0+c_2\cos 0)$

$2=4\sqrt{6}{c}_2$

$c_2=\frac{1}{2\sqrt{6}}$

Substitute$c_1=0,c_2=\frac{1}{2\sqrt{6}}$ values in the equation$x\left(t\right)$

$x\left(t\right)=0\cdot \cos 4\sqrt{6}t+\frac{1}{2\sqrt{6}}\sin 4\sqrt{6}t$

$x\left(t\right)=\frac{1}{2\sqrt{6}}\sin 4\sqrt{6}t$

Therefore, the required equation of motion is, $x\left(t\right)=\frac{\sqrt{6}}{12}\sin 4\sqrt{6}t$.