I need help solving this problem. We must show all work. A mass streches a sprin

Question

I need help solving this problem. We must show all work.

A mass streches a spring 2.45m whose spring constant is 8N/m. The mass is intially released from the equilibrium position with a downward velocity of 1/25 m/s, and the subsequent motion takes motion takes place in a medium that offers a damping force that is numerically equal to 8 times the instantaneously velocity. Find the equation of the motion if the mass is driven by an external force equal to f(t) = 2e-tcos2

Identify the following for the associated homgenous DE

auxiliary equation:__________ ; Find both roots of the auxiliary equation________________

Find both solutions of the auxiliary equation: Xc1(t)=____________& Xc2(t)=_______________

Give the complementary function to the DE above (use C1 and C2 for the undetermined coefficients):

Xc(t)=________________________________________________________________________

Identify the following when using the annihilator approach: F(t)=__________________________

Which differential operator is used to annihilate F(t)?

Give the aprticular solution to the nonhomgenous DE (use A and B for the undetermined coefficients):

Xp(t)=__________________________________________________________________________

To find the undetermined coefficents A and B of the particular solution you will also need:

X/p(t)=___________________________________________________________________________

X//p(t)=___________________________________________________________________________

Write the general solution to the equation of motion, x(t)=xc(t)+xp(p), as a function of time,t.

to find the undetermined coefficients C1 and C2 with the intial conditions given , you will aslo need:

X/(t)=____________________________________________________________________________________

The Equation of Motion is x(t) = ________________________________________________________________

Explanation / Answer

First we solve for the mass of the object. We know that the spring has a proportional force related to displacement. Since the object stretches it 2.45m, we can use hooks law to determine the following:

F = 8N/m * 2.45m = 19.6N = mg = m(9.8)

Therefore the mass equals 2kg.

We set up a force equation:

Force = mass times acceleration, from which we get the term 2x''. We want to add to that a force equal to 8 times the velocity from which we get the term 8x'. And finally we want to add the force from the spring constant which is equal to 8 N/m times the displacement from which we get the term 8x.

Therefore our auxiliary equation, not including the driving force is:

2x''+8x'+8x = 0

We divide by 2 to get our equation in standard linear form:

x''+4x'+4x=0

To find both roots we guess the root emt. Differentiating emt twice gives us the following:

x = emt, x'=memt, x''=m2emt

We then substitute these into our original equation which gives:

x=emt(m2+4m+4)=0

Since the exponential cannot equal zero, we know:

m2+4m+4=0

(m+2)2 = 0

We only get one root from this: m=-2, so we have the first solution of the auxillery equation:

xc1=C1e-2t

To find the 2nd solution, we multiply this through by t:

xc2=C2te-2t

So our complementery solution is:

xc= C1e-2t + C2te-2t

Now the problem wants us to find a differential annihilator to eliminate F(t)=2e-tcos(2t)

The annihilator that gets rid of eaxcos(Bx) is (D2-2aD+(a2+B2))

Remember if F(t) had been in the proper position in the original ODE, the two would have been divided out, so we are really working with the function: F(t)=e-tcos(2t)

So, our anihillator operator is: (D2+2D+5)

We equate this to zero to find the roots of our particular solution:

D2+2D+5=0

D2+2D = -5

(D+1)2=-4

D = -1 +/- 2i

Which gives the following particular solution:

xp = e-t( Acos(2t) + Bsin(2t) )

Now we find the first and second derivative of the particular solution:

x'p = - e-t( Acos(2t) + Bsin(2t) ) + e-t( -2Asin(2t) + 2Bcos(2t) )

x''p = e-t( Acos(2t) + Bsin(2t) ) - e-t( -2Asin(2t) + 2Bcos(2t) ) - e-t( -2Asin(2t) + 2Bcos(2t) ) + e-t( -4Acos(2t) - 4Bsin(2t) )

Now, to find A and B, we substitute these into our general equation x''+4x'+4=e-tcos(2t)

e-t( Acos(2t) + Bsin(2t) ) - e-t( -2Asin(2t) + 2Bcos(2t) ) - e-t( -2Asin(2t) + 2Bcos(2t) ) + e-t( -4Acos(2t) - 4Bsin(2t) ) - e-t( 4Acos(2t) + 4Bsin(2t) ) + e-t( -8Asin(2t) + 8Bcos(2t) ) + e-t( 4Acos(2t) + 4Bsin(2t) ) = e-tcos(2t)

Pull out e^-t:

e-t( Acos(2t) + Bsin(2t) + 2Asin(2t) - 2Bcos(2t) + 2Asin(2t) - 2Bcos(2t) - 4Acos(2t) - 4Bsin(2t) - 4Acos(2t) - 4Bsin(2t) - 8Asin(2t) + 8Bcos(2t) + 4Acos(2t) + 4Bsin(2t) ) = e-tcos(2t)

Group Cosines and Sines, and A's and B's:

e-t( Acos(2t) - 4Acos(2t) - 4Acos(2t) + 4Acos(2t) - 2Bcos(2t) - 2Bcos(2t) + 8Bcos(2t) + 2Asin(2t) + 2Asin(2t) - 8Asin(2t) + Bsin(2t) - 4Bsin(2t) - 4Bsin(2t)   + 4Bsin(2t) ) = e-tcos(2t)

Simplify:

e-t( -3Acos(2t) + 4Bcos(2t) - 4Asin(2t) - 3Bsin(2t) ) = e-tcos(2t)

Now we have the general set of equations:

-3A+4B = 1

-4A-3B = 0

Therefore

A= -3/25

B= 4/25

Now our general solution is:

xg= C1e-2t + C2te-2t + e-t( -.12cos(2t) + .16sin(2t) )

Now we solve for our first initial condition, x(0)=0

Which gives this equation:

0 = C1 - .12

C1 = .12

Substitute in to the original equation and take the derivative to solve for the 2nd initial condition x'(0)=1/25

xg= .12e-2t + C2te-2t + e-t( -.12cos(2t) + .16sin(2t) )

x'g= -.24e-2t -2C2te-2t + C2e-2t - e-t( -.12cos(2t) + .16sin(2t) ) + e-t( .24sin(2t) + .32cos(2t) )

x'g(0)= -.24 + C2 - .12 + .32 = 1/25

C2=1/25

Now we have the final equation of motion:

xg= .12e-2t + .04(t)e-2t + e-t( -.12cos(2t) + .16sin(2t) )

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