Respuesta :
Answer:
[tex]x(t)=e^{-t}cos2t+3sin2t[/tex]
Explanation:
We are given that
Mass,m=1 slug
Spring constant,k=5lb/ft
x(0)=1 foot
Velocity,x'(0)=5ft/s
[tex]\beta=2[/tex]
[tex]f(t)=12cos2t+3sin2t[/tex]
[tex]\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=\frac{1}{m}f(t)[/tex]
Substitute the values
[tex]\frac{d^2x}{dt^2}+2\frac{dx}{dt}+5x=12cos2t+3sin2t[/tex]
Auxillary equation
[tex]m^2+2m+5=0[/tex]
[tex]m=\frac{-2\pm \sqrt{4-4\times 5}}{2}=\frac{-2\pm 4i}{2}[/tex]
By using quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]m=\frac{-2+4i}{2}=-1+2i[/tex]
[tex]m=\frac{-2-4i}{2}=-1-2i[/tex]
Complementary solution
[tex]y_c(x)=e^{-t}(Acos2t+Bsin2t)[/tex]
[tex]x_p=acos2t+bsin2t[/tex]
[tex]x'_p=-2asin2t+2bcos2t[/tex]
[tex]x''_p=-4acos2t-4bsin2t[/tex]
Substitute the values
[tex]-4acos2t-4bsin2t+-4asin2t+4bcos2t+5acos2t+5bsin2t=12cos2t+3sin2t[/tex]
[tex]acos2t+4bcos2t+bsin2t-4asin2t=12cos2t+3sin2t[/tex]
[tex](a+4b)cos2t+(b-4a)sin2t=12cos2t+3sin2t[/tex]
Comparing coefficient on both sides
[tex]a+4b=12[/tex]....(1)
[tex]-4a+b=3[/tex]...(2)
Equation (1) multiply by 4 and then add to equation (2)
[tex]17b=51[/tex]
[tex]b=\frac{51}{17}=3[/tex]
Substitute the value of b in equation (2)
[tex]-4a+3=3[/tex]
[tex]-4a=3-3=0[/tex]
[tex]a=0[/tex]
Therefore, [tex]x_p=3sin2t[/tex]
General solution ,[tex]x(t)=x_c(t)+x_p(t)=e^{-t}(Acos2t+Bsin2t)+3sin2t[/tex]
[tex]x(0)=A[/tex]
[tex]1=A[/tex]
[tex]x'(t)=-e^{-t}(Acos2t+Bsin2t)+e^{-t}(-2Asin2t+2Bcos2t)+6cos2t[/tex]
[tex]x'(0)=-A+2B+6[/tex]
[tex]5=-1+2B+6=5+2B[/tex]
[tex]2B=5-5=0[/tex]
[tex]B=0[/tex]
Substitute the values
[tex]x(t)=e^{-t}cos2t+3sin2t[/tex]
The equation of motion if the mass is driven by an external force is ;
x(t) = [tex]e^{-t} cos2t + 3sin2t[/tex]
Given data :
mass ( m ) = 1 slug
constant ( k ) = 5 Ib/ft
Initial velocity ( x'(0) ) = 5 ft/s
Initial height ( x ) = 1 ft
Damping force ( β ) = 2
Determine the equation of motion if the mass is driven by an external force
f(t) = [tex]12cos2t + 3sin2t[/tex]
[tex]\frac{d^2x}{dt^2} + \frac{\beta }{m} \frac{dx}{dt} + \frac{k}{m}x = \frac{1}{m}f(t)[/tex] ( second derivative ) -------- ( 2 )
Insert the given values into equation ( 2 )
The auxiliary equation
[tex]m^{2} + 2m + 5 = 0[/tex]
m = [tex]\frac{-2i+4i}{2}[/tex] applying quadratic formula
m = -1 + 2i , and - 1 - 2i
Expressing the complementary solutions
[tex]x_{p} = acos2t + bsin2t \\x'_{p} = -2asin2t + 2bcos2t \\x"_{p} = -4acos2t - 4bsin2t[/tex]
substituting these values
we will get ;
( a + 4b)cos2t + ( b - 4a )sin2t = 12cos2t + 3sin2t
comparing the coefficients on both sides
a + 4b = 12 ----- ( 3 )
- 4a + b = 3 --- ( 4 )
resolving equations 3 and 4 simultaneously
a = 0
b = 3
∴ [tex]x_{p}[/tex] = 3sin2t
x'(0) = -A + 2B + 6
5 = -1 + 2B + 6 = 5 + 2B
∴ B = 0
Hence we can conclude that the equation of motion if the mass is driven by an external force is
x(t) = [tex]e^{-t} cos2t + 3sin2t[/tex]
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