Question

A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of 5 ft/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 2 times the instantaneous velocity. (a) Find the equation of motion if the mass is driven by an external force equal to f(t)

Answer 1

Answer:

$x(t)=e^{-t}cos2t+3sin2t$

Explanation:

We are given that

Mass,m=1 slug

Spring constant,k=5lb/ft

x(0)=1 foot

Velocity,x'(0)=5ft/s

$\beta=2$

$f(t)=12cos2t+3sin2t$

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=\frac{1}{m}f(t)$

Substitute the values

$\frac{d^2x}{dt^2}+2\frac{dx}{dt}+5x=12cos2t+3sin2t$

Auxillary equation

$m^2+2m+5=0$

$m=\frac{-2\pm \sqrt{4-4\times 5}}{2}=\frac{-2\pm 4i}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$m=\frac{-2+4i}{2}=-1+2i$

$m=\frac{-2-4i}{2}=-1-2i$

Complementary solution

$y_c(x)=e^{-t}(Acos2t+Bsin2t)$

$x_p=acos2t+bsin2t$

$x'_p=-2asin2t+2bcos2t$

$x''_p=-4acos2t-4bsin2t$

Substitute the values

$-4acos2t-4bsin2t+-4asin2t+4bcos2t+5acos2t+5bsin2t=12cos2t+3sin2t$

$acos2t+4bcos2t+bsin2t-4asin2t=12cos2t+3sin2t$

$(a+4b)cos2t+(b-4a)sin2t=12cos2t+3sin2t$

Comparing coefficient on both sides

$a+4b=12$....(1)

$-4a+b=3$...(2)

Equation (1) multiply by 4 and then add to equation (2)

$17b=51$

$b=\frac{51}{17}=3$

Substitute the value of b in equation (2)

$-4a+3=3$

$-4a=3-3=0$

$a=0$

Therefore, $x_p=3sin2t$

General solution ,$x(t)=x_c(t)+x_p(t)=e^{-t}(Acos2t+Bsin2t)+3sin2t$

$x(0)=A$

$1=A$

$x'(t)=-e^{-t}(Acos2t+Bsin2t)+e^{-t}(-2Asin2t+2Bcos2t)+6cos2t$

$x'(0)=-A+2B+6$

$5=-1+2B+6=5+2B$

$2B=5-5=0$

$B=0$

Substitute the values

$x(t)=e^{-t}cos2t+3sin2t$