Question

2. < Previous A mass that weight 6 lb stretches a spring 2 in. The system is acted on by an external force 5 sin ( 8V3t) lb. If the mass is pushed up 1 in and then released, determine the position of the mass at any time t. Use 32 ft/s’ as the acceleration due to gravity. Pay close attention to the units. u(t) = in

Answer 1

weight is w=6 pound

gravity is 32 m/s²

mass is given by

$m=\frac{w}{g}$

6 m 32

$m=\frac{3}{16}$ slug

.

a force of 6 pounds, stretches a spring 2inch = 1/6 feet

so x=1/6

from the Hooke's law, spring constant k is

$F=kx$

$6=k\left(\frac{1}{6}\right)$

$k=36$

.

.

there is no damping, so damping constant is $c=0$

.

external force is $f\left(t\right)=5\sin \left(8\sqrt{3}t\right)$

DE is given by

my' + cy + ky = f(t)

3 16 ! +0y + 36y = 5 sin (8V3t

$\frac{3}{16}y''+36y=5\sin \left(8\sqrt{3}t\right)$

$3y''+576y=80\sin \left(8\sqrt{3}t\right)$

find roots for a homogeneous system

$3r^2+576=0$

$3r^2=-576$

$r^2=-192$

$r=\sqrt{-192}$

$r= \pm 8\sqrt{3}i$

for complex roots, the complementary solution is

$y_c=e^0\left(c_1\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right)\right)$

$y_c= c_1\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right)$....................(1)

.

on the right of the DE, we have

$g\left(t\right)=80\sin \left(8\sqrt{3}t\right)$

so assumed that the particular solution is

$y=a_0t\sin \left(8\sqrt{3}t\right)+a_1t\cos \left(8\sqrt{3}t\right)$.................(1)

$y'=a_0\left(\sin \left(8\sqrt{3}t\right)+8\sqrt{3}t\cos \left(8\sqrt{3}t\right)\right)+a_1\left(\cos \left(8\sqrt{3}t\right)-8\sqrt{3}t\sin \left(8\sqrt{3}t\right)\right)$

$y''=a_0\left(16\sqrt{3}\cos \left(8\sqrt{3}t\right)-192t\sin \left(8\sqrt{3}t\right)\right)+a_1\left(-16\sqrt{3}\sin \left(8\sqrt{3}t\right)-192t\cos \left(8\sqrt{3}t\right)\right)$

.

put all values in DE

$3y''+576y=80\sin \left(8\sqrt{3}t\right)$

${\color{black} 3\left(a_0\left(16\sqrt{3}\cos \left(8\sqrt{3}t\right)-192t\sin \left(8\sqrt{3}t\right)\right)+a_1\left(-16\sqrt{3}\sin \left(8\sqrt{3}t\right)-192t\cos \left(8\sqrt{3}t\right)\right)\right)+576\left(a_0t\sin \left(8\sqrt{3}t\right)+a_1t\cos \left(8\sqrt{3}t\right)\right)=80\sin \left(8\sqrt{3}t\right)}$

${\color{black} 48\sqrt{3}a_0\cos \left(8\sqrt{3}t\right)-576a_0t\sin \left(8\sqrt{3}t\right)-48\sqrt{3}a_1\sin \left(8\sqrt{3}t\right)-576a_1t\cos \left(8\sqrt{3}t\right)+576a_0t\sin \left(8\sqrt{3}t\right)+576a_1t\cos \left(8\sqrt{3}t\right)=80\sin \left(8\sqrt{3}t\right)}$

$48\sqrt{3}a_0\cos \left(8\sqrt{3}t\right)-48\sqrt{3}a_1\sin \left(8\sqrt{3}t\right)=80\sin \left(8\sqrt{3}t\right)$

compare Coefficient both side

$\begin{bmatrix}80=-48\sqrt{3}a_1\\ 0=48\sqrt{3}a_0\end{bmatrix}$

$a_0=0,\:a_1=-\frac{5\sqrt{3}}{9}$...............put it back in equation 1

.

$y=a_0t\sin \left(8\sqrt{3}t\right)+a_1t\cos \left(8\sqrt{3}t\right)$

$y=0\cdot \:t\sin \left(8\sqrt{3}t\right)+\left(-\frac{5\sqrt{3}}{9}\right)t\cos \left(8\sqrt{3}t\right)$

$y= -\frac{5\sqrt{3}}{9}t\cos \left(8\sqrt{3}t\right)$

$y_p= -\frac{5\sqrt{3}}{9}t\cos \left(8\sqrt{3}t\right)$

.

.

general solution is

y = y + yp

$y=c_1\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right) -\frac{5\sqrt{3}}{9}t\cos \left(8\sqrt{3}t\right)$................(3)

here mass is initially released from rest from 1 inch above the equilibrium position.

so y(0)=-1

$-1=c_1\cos \left(8\sqrt{3}\cdot \:0\right)+c_2\sin \left(8\sqrt{3}\cdot \:0\right)-\frac{5\cdot \:0\cdot \cos \left(8\sqrt{3}\cdot \:0\right)}{3^{\frac{3}{2}}}$

$c_1=-1$..............put it back in equation 3

.

$y=c_1\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right) -\frac{5\sqrt{3}}{9}t\cos \left(8\sqrt{3}t\right)$................(4)

$y=-\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right)-\frac{5t\cos \left(8\sqrt{3}t\right)}{3\sqrt{3}}$

take derivative

$y'=8\sqrt{3}\sin \left(8\sqrt{3}t\right)+8\sqrt{3}c_2\cos \left(8\sqrt{3}t\right)-\frac{5}{3\sqrt{3}}\cos \left(8\sqrt{3}t\right)+\frac{40t\sin \left(8\sqrt{3}t\right)}{3}$

here initial velocity is zero so y'(0)=0$0=8\sqrt{3}\sin \left(8\sqrt{3}\cdot \:0\right)+8\sqrt{3}c_2\cos \left(8\sqrt{3}\cdot \:0\right)-\frac{5}{3^{\frac{3}{2}}}\cos \left(8\sqrt{3}\cdot \:0\right)+\frac{40\cdot \:0\cdot \sin \left(8\sqrt{3}\cdot \:0\right)}{3}$

$0= 8\sqrt{3}c_2-\frac{5\sqrt{3}}{9}$

$8\sqrt{3}c_2=\frac{5\sqrt{3}}{9}$

$c_2=\frac{5}{72}$..............put it back in the equation 4

$y=c_1\cos \left(8\sqrt{3}t\right)+c_2\sin \left(8\sqrt{3}t\right) -\frac{5\sqrt{3}}{9}t\cos \left(8\sqrt{3}t\right)$

$y=-\cos \left(8\sqrt{3}t\right)+\frac{5}{72}\sin \left(8\sqrt{3}t\right)-\frac{5t\cos \left(8\sqrt{3}t\right)}{3\sqrt{3}}$

${\color{Red} u(t)=-\cos \left(8\sqrt{3}t\right)+\frac{5}{72}\sin \left(8\sqrt{3}t\right)-\frac{5t\cos \left(8\sqrt{3}t\right)}{3\sqrt{3}}}$