The differential equation that governs the forced oscillation is shown below:
0.2 d²y/dt² + 1.2 dy/dt +2y = r(t) where r(t) is the external force.
Given that r(t) = 5 cos 4t with y(0) = 0 . find the equation of motion of the forced oscillations
Normalize the equation by multiplying by 5:
0.2 d²y/dt² + 1.2 dy/dt +2y = 5 cos(4t) = r(t)
to:
d²y/dt² + 6 dy/dt + 10y = 25cos(4t)
Find the complementary solution:
m²+6m+10=0
m=-3±i
So the solution to the homogeneous equation is:
yc=e^(-3t)(C1*cos(t)+C2*sin(t))
Now find the particular solution by undetermined coefficients:
Assume the particular solution to be:
yp=Acos(4t)+Bsin(4t)
and substitute in y of the the original equation:
d²yp/dt² + 6 dyp/dt + 10yp = 25cos(4t)
-16Acos(4t)-16Bsin(4t)
+6(4Bcos(4t)-4Asin(4t))
+10Acos(4t)+10Bsin(4t)
=(-6A+24B)cos(4t)+(-24A-6B)sin(4t)
Compare coefficients of cos(4t) and sin(4t):
-24A-6B=0 => B=-4A
-6A+24B=25 => -102A=25 => A=-25/102
Therefore
yp(t)=-(25/102)cos(4t)+(100/102)sin(4t)
(substitute in homogeneous equation to verify that you get 25cos(4t) )
The general solution is therefore:
y=yc+yp=e^(-3t)(C1*cos(t)+C2*sin(t))-(25/102)cos(4t)+(100/102)sin(4t)
Initial conditions:
To solve the second order problem completely, you'll need two initial conditions. We are givn y(0)=0 at t=0.
We need another one (such as y'(0)=5 at t=0).
Substitute the initial conditions into the general solution above and solve for C1 and C2 to give the final solution of the initial value problem.
To find the equation of motion for forced oscillations given the differential equation and the external force, follow these steps:
Step 1: Rewrite the given differential equation using standard notation:
0.2y'' + 1.2y' + 2y = r(t)
Step 2: Substitute the given external force r(t) = 5cos(4t) into the differential equation:
0.2y'' + 1.2y' + 2y = 5cos(4t)
Step 3: Solve the homogeneous equation by setting r(t) = 0:
0.2y'' + 1.2y' + 2y = 0
Step 4: Solve the homogeneous equation by assuming a solution of the form y = e^(mt):
0.2(m^2)e^(mt) + 1.2me^(mt) + 2e^(mt) = 0
Step 5: Divide each term by e^(mt):
0.2m^2 + 1.2m + 2 = 0
Step 6: Solve the quadratic equation to find the values of m:
m^2 + 6m + 10 = 0
Using the quadratic formula, we get:
m = (-6±√(6^2-4*1*10))/(2*1)
m = (-6±√(-4))/2
m = (-6±2i)/2
m = -3±i
Step 7: Write the general solution for the homogeneous equation using the roots found in step 6:
y_h(t) = C1e^(-3t)cos(t) + C2e^(-3t)sin(t)
Step 8: Use the method of undetermined coefficients to find a particular solution for the non-homogeneous equation when r(t) = 5cos(4t):
Assume a particular solution of the form y_p(t) = Acos(4t) + Bsin(4t), where A and B are constants to be determined.
Differentiate y_p(t) twice to find the necessary derivatives:
y_p'(t) = -4Asin(4t) + 4Bcos(4t)
y_p''(t) = -16Acos(4t) - 16Bsin(4t)
Substitute these derivatives into the original differential equation:
0.2(-16Acos(4t) - 16Bsin(4t)) + 1.2(-4Asin(4t) + 4Bcos(4t)) + 2(Acos(4t) + Bsin(4t)) = 5cos(4t)
Simplify the equation and collect terms:
(-3.2A + 4B)cos(4t) + (-4.8A - 3.2B)sin(4t) = 5cos(4t)
Equating coefficients, we get two equations:
-3.2A + 4B = 5 (for the cosine term)
-4.8A - 3.2B = 0 (for the sine term)
Solving these equations simultaneously, we find:
A = -1.25
B = -1.56
Therefore, the particular solution is:
y_p(t) = -1.25cos(4t) - 1.56sin(4t)
Step 9: Write the complete solution for the non-homogeneous equation by combining the homogeneous and particular solutions:
y(t) = y_h(t) + y_p(t)
= C1e^(-3t)cos(t) + C2e^(-3t)sin(t) - 1.25cos(4t) - 1.56sin(4t)
Given that y(0) = 0, substitute t = 0 into the equation:
0 = C1e^(0)cos(0) + C2e^(0)sin(0) - 1.25cos(0) - 1.56sin(0)
0 = C1 - 1.25
Therefore, C1 = 1.25.
Step 10: Write the final equation of motion for the forced oscillations:
y(t) = 1.25e^(-3t)cos(t) + C2e^(-3t)sin(t) - 1.25cos(4t) - 1.56sin(4t)