Show all work necessary to arrive at your answers. A mass stretches a spring 2.4
ID: 2942623 • Letter: S
Question
Show all work necessary to arrive at your answers.
A mass stretches a spring 2.45-m whose spring constant is 8-N/m. The mass is initially released from the equilibrium position with a downward velocity of 1/25 m/s, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 8 times the instantaneous velocity.
Find the equation of motion if the mass is driven by an external force equal to:
f(t)=2e-tcos2t
Write the correct Driven Motion Equation.
Write the above Differential Equation in Differential Operator Form.
Give the complementary function to the above DE xc(t)= ?
Give the particular solution to the Nonhomogeneous DE (use A and B for the undetermined coefficients)
Write the General Solution to the Equation of Motion
Explanation / Answer
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) )