Physics, 30.03.2020 19:26 lilpeepxliltracy

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)

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Answer from: katwright1124

The equation of motion if the mass is driven by an external force is ;

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

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) = 12cos2t + 3sin2t

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

Insert the given values into equation ( 2 )

The auxiliary equation

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

m = $\frac{-2i+4i}{2}$ applying quadratic formula

m = -1 + 2i , and - 1 - 2i

Expressing the complementary solutions

$x_{p} = acos2t + bsin2t \\x'_{p} = -2asin2t + 2bcos2t \\x"_{p} = -4acos2t - 4bsin2t$

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

∴ $x_{p}$ = 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) = $e^{-t} cos2t + 3sin2t$

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Answer from: ilovebeanieboos

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

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

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

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A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially, the mass is released...

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