Answer :
Answer:
The answer is
"[tex]x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))[/tex]".
Explanation:
Taking into consideration a volume weight = 16 pounds originally extends a springs [tex]\frac{8}{3}[/tex] feet but is extracted to resting at 2 feet beneath balance position.
The mass value is =
[tex]W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\[/tex]
The source of the hooks law is stable,
[tex]16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\[/tex]
Number [tex]\frac{1}{2}[/tex] times the immediate speed, i.e .. Damping force
[tex]\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\[/tex]
The m^2+m+12=0 and m is an auxiliary equation,
[tex]m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}[/tex]
Therefore, additional feature
[tex]x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t][/tex]
Use the form of uncertain coefficients to find a particular solution.
Assume that solution equation,
[tex]x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\[/tex]
These values are replaced by equation ( 1):
[tex]\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\[/tex]
Going to compare cos3 t and sin 3 t coefficients from both sides,
The cost3 t is 3A + 3B= 20 coefficients
The sin 3 t is 3B -3A = 0 coefficient
The two equations solved:
[tex]3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\[/tex]
Replace the very first equation with the meaning,
[tex]3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\[/tex]
equation is
[tex]x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)[/tex]
The ultimate plan for both the equation is therefore
[tex]x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)[/tex]
Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.
Throughout the general solution, replace initial state x(0) = 2,
Replace x'(0)=0 with a general solution in the initial condition,
[tex]x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\[/tex]
[tex]x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\[/tex]
[tex]c_2=\frac{-64\sqrt{47}}{141}[/tex]
[tex]x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))[/tex]