Respuesta :
Answer with Step-by-step explanation:
Let a mass weighing 16 pounds stretches a spring [tex]\frac{8}{3}[/tex] feet.
Mass=[tex]m=\frac{W}{g}[/tex]
Mass=[tex]m=\frac{16}{32}[/tex]
[tex]g=32 ft/s^2[/tex]
Mass,m=[tex]\frac{1}{2}[/tex] Slug
By hook's law
[tex]w=kx[/tex]
[tex]16=\frac{8}{3} k[/tex]
[tex]k=\frac{16\times 3}{8}=6 lb/ft[/tex]
[tex]f(t)=10cos(3t)[/tex]
A damping force is numerically equal to 1/2 the instantaneous velocity
[tex]\beta=\frac{1}{2}[/tex]
Equation of motion :
[tex]m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)[/tex]
Using this equation
[tex]\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)[/tex]
[tex]\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)[/tex]
[tex]\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)[/tex]
Auxillary equation
[tex]m^2+m+12=0[/tex]
[tex]m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}[/tex]
[tex]m=\frac{-1\pmi\sqrt{47}}{2}[/tex]
[tex]m_1=\frac{-1+i\sqrt{47}}{2}[/tex]
[tex]m_2=\frac{-1-i\sqrt{47}}{2}[/tex]
Complementary function
[tex]e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})[/tex]
To find the particular solution using undetermined coefficient method
[tex]x_p(t)=Acos(3t)+Bsin(3t)[/tex]
[tex]x'_p(t)=-3Asin(3t)+3Bcos(3t)[/tex]
[tex]x''_p(t)=-9Acos(3t)-9sin(3t)[/tex]
This solution satisfied the equation therefore, substitute the values in the differential equation
[tex]-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)[/tex]
[tex](3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)[/tex]
Comparing on both sides
[tex]3B+3A=20[/tex]
[tex]3B-3A=0[/tex]
Adding both equation then, we get
[tex]6B=20[/tex]
[tex]B=\frac{20}{6}=\frac{10}{3}[/tex]
Substitute the value of B in any equation
[tex]3A+10=20[/tex]
[tex]3A=20-10=10[/tex]
[tex]A=\frac{10}{3}[/tex]
Particular solution, [tex]x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)[/tex]
Now, the general solution
[tex]x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)[/tex]
From initial condition
x(0)=2 ft
x'(0)=0
Substitute the values t=0 and x(0)=2
[tex]2=c_1+\frac{10}{3}[/tex]
[tex]2-\frac{10}{3}=c_1[/tex]
[tex]c_1=\frac{-4}{3}[/tex]
[tex]x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)[/tex]
Substitute x'(0)=0
[tex]0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2[/tex]
[tex]\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0[/tex]
[tex]\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}[/tex]
[tex]c_2==-\frac{64}{3\sqrt{47}}[/tex]
Substitute the values then we get
[tex]x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)[/tex]
