Question

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the equilibrium position.

(a) Find the position of the mass at the times $\mathbf{t}\mathbf{=}\mathbf{\pi }\mathbf{/}\mathbf{12}\mathbf{,}\mathbf{\pi }\mathbf{/}\mathbf{8}$, $\mathbf{\pi }\mathbf{/}\mathbf{6}\mathbf{,}\mathbf{\pi }\mathbf{/}\mathbf{4}$, and $\frac{9\pi }{32}\mathbf{}\mathbf{s}$.

(b) What is the velocity of the mass when $\mathbf{t}\mathbf{=}\frac{3\pi }{16}\mathbf{s}$? In which direction is the mass heading at this instant?

(c) At what times does the mass pass through the equilibrium position?

Short answer

(a)So, the position of the mass are$3,6,3,-6,-3\sqrt{2}$.

(b)So, the velocity of the mass$48\text{inch}/\text{s}$.

(c)So, the required time is $t=\frac{n\pi }{8}$.

Step-by-step solution

Definition of Newton’s second law

The second law of motion describes what happens to the massive body when acted upon by an external force.

(a)Step 2: Use of Newton's second law

Consider the spring of 6 inches length stretched by a mass of 20 pounds. It is known that initially, the mass is released from a resting point 6 inches below the equilibrium position.

Objective is to determine the position of the mass at times .

$t=\pi /12,\pi /8,\pi /6,\pi /4,9\pi /32$

According to the Newton's second law,

role="math" localid="1667911781923" $m\frac{d^{2}x}{d{t}^2}=-kx$

Or,

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

where $m$is the mass,$x$ is the displacement and $k$is the spring constant.

Find the Mass

The auxiliary equation of this homogeneous differential equation is:

$r^2+\frac{k}{m}=0$

And then, $r=\sqrt{\frac{k}{m}}i,-\sqrt{\frac{k}{m}}i$.

Then the general solution is:$x\left(t\right)=c_1\cos \left(\sqrt{\frac{k}{m}}t\right)+c_2\sin \left(\sqrt{\frac{k}{m}}t\right)$

The mass is

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

Find the length

The length (in feet) is$s=\frac{6}{12}$ , and

$\begin{array}{rcl}k& =& \frac{W}{s}\\ & =& \frac{20}{1/2}\\ & =& 40lb/f\end{array}$

Then the displacement will be:

$x\left(t\right)=c_1\cos \left(\sqrt{\frac{40}{5/8}}t\right)+c_2\sin \left(\sqrt{\frac{40}{5/8}}t\right)$

$=c_1\cos \left(8t\right)+c_2\sin \left(8t\right)$

$x^{\text{'}}\left(t\right)=-8c_1\sin 8t+8c_2\cos 8t$

Use initial condition

According to the known information, the initial conditions are:

$x\left(0\right)=-6\mathrm{in}\left(=-\frac{1}{2}\mathrm{ft}\right),\frac{dx\left(0\right)}{dt}=0ft/s.$

Then

$-\frac{1}{2}=c_1\cos 0+c_2\sin 0$

$c_1=-\frac{1}{2}$

$0=-8c_1\sin 0+8c_2\cos 0$

$c_2=0$

Thus,$x\left(t\right)=-\frac{1}{2}\cos \left(8t\right)ft$ Or,$x\left(t\right)=-6\cos \left(8t\right)$ inch.

Hence,

$\begin{array}{rcl}x\left(\frac{\pi }{12}\right)& =& -6\cos \left(\frac{8\pi }{12}\right)\\ & =& 3\end{array}$

role="math" localid="1667912713181" $\begin{array}{rcl}x\left(\frac{\pi }{8}\right)& =& -6\cos \left(\frac{8\pi }{8}\right)\\ & =& 6\end{array}$

$\begin{array}{rcl}x\left(\frac{\pi }{6}\right)& =& -6\cos \left(\frac{8\pi }{8}\right)\\ & =& 3\end{array}$

$x\left(\frac{\pi }{4}\right)=-6$

$x\left(\frac{9\pi }{32}\right)=-3\sqrt{2}$

(b)Step 1: Find the mass velocity

Objective is to determine the mass velocity at $t=\frac{3\pi }{16}$seconds along with its direction.

The velocity of mass is given by $\dot{x}\left(t\right)$.

Since $x\left(t\right)=-6\cos \left(8t\right)$inch, therefore

$\begin{array}{rcl}\dot{x}\left(t\right)& =& -6\frac{d}{dt}\left(\cos 8t\right)\\ & =& 48\sin 8t\end{array}$

At time $t=\frac{3\pi }{16}$,

$\dot{x}\left(\frac{3\pi }{16}\right)=48\sin \left(8\cdot \frac{3\pi }{16}\right)$

$=48\sin \left(\frac{3\pi }{2}\right)$

$=-48\text{inch/s}$

Negative sign tells the downward direction.

That is, mass moves downward with the velocity $48\text{inch}/\text{s}$.

(C)Step 1: Find the time

Objective is to determine the time when mass pass through the equilibrium position.

In the equilibrium position,$\dot{x}\left(t\right)=0$ .

That is, $\sin 8t=0$.

And

$\sin 8t=\sin n\pi$

$8t=n\pi$

$t=\frac{n\pi }{8},(n=0,1,2,\dots )$

Hence, at time$t=\frac{n\pi }{8}$ second the mass will pass through the equilibrium position.