Differential equations, initial value problem.
The general equation of motion is:
mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable.
In this case, external force f(t)=0, so
mx"+Bx'+kx=0
substitute
m=0.25,
k=4,
B=1
we have a differential equation of motion with constant coefficients:
0.25x"+x'+4x=0
Divide by the leading coefficient:
x"+4x'+16x=0
Auxiliary equation:
m²+4m+16=0
m1=-2+2√3, m2=-2-2√3
α=-2, β=2√3
So the solution for x is:
x=e-2t(C1*cos(βt)+C2*sin(βt))
Since x(0)=-0.5, we have
-0.5=1*(C1+C2*0)
or C1=-0.5
Find
x'
=dx/dt
=a*%e^(a*t)*(sin(b*t)*C2+cos(b*t)*C1)+%e^(a*t)*(b*cos(b*t)*C2-b*sin(b*t)*C1)
substitute
x'(0)=-1 to get
-1=αC1+βC2
-1=-2C1+2√(3)C2
C2=(-1+2(-0.5))/2√(3)
=-√(3)/3
Therefore the equation of motion is:
x(t)=e-2t(-0.5cos(βt)-(√(3)/3)sin(βt))
Is this a question on the above response?
(see: http://www.jiskha.com/display.cgi?id=1299811520)
The solution to the differential equation is:
x(t) = e^(-2t)(-0.5cos(2√3t) - (√3/3)sin(2√3t))
To find x'(t), the derivative of x with respect to t:
x'(t) = e^(-2t)(-0.5[2√3sin(2√3t)] - (√3/3)[2√3cos(2√3t)])
= e^(-2t)(-√3sin(2√3t) - 2cos(2√3t))
And finally, substituting x'(0) = -1:
-1 = -2C1 + 2√3C2
Solving for C2:
C2 = (-1 + 2C1)/2√3
= (-1 + 2(-0.5))/2√3
= -√3/3
Therefore, the equation of motion is:
x(t) = e^(-2t)(-0.5cos(2√3t) - (√3/3)sin(2√3t))
To get the answer to the question, we need to solve the given differential equation using initial value conditions. The general equation of motion is mx" + Bx' + kx = f(t), where m, B, k are constants and f(t) is the external force acting on the system.
In this case, the external force f(t) is given as 0, so the equation becomes mx" + Bx' + kx = 0. We are given the values of m = 0.25, B = 1, and k = 4. Dividing the entire equation by the leading coefficient, we get x" + 4x' + 16x = 0.
To solve this differential equation, we need to find the roots of the auxiliary equation m^2 + 4m + 16 = 0. Using the quadratic formula, we find the roots as m₁ = -2 + 2√3 and m₂ = -2 - 2√3.
The general solution for x is given by x = e^(mt)(C₁cos(βt) + C₂sin(βt)), where β = √(4ac - b²)/2a. In this case, α and β are the roots of the auxiliary equation, and a = 1, b = 4, and c = 16. So, β = √(4(1)(16) - (4)²)/2(1) = 2√3.
Now, we need to find the values of C₁ and C₂ using the initial value condition x(0) = -0.5. Substituting t = 0 and x = -0.5 into the general solution, we get -0.5 = e^(m(0))(C₁cos(β(0)) + C₂sin(β(0))). Simplifying this equation, we find C₁ = -0.5.
To find C₂, we need to differentiate the general solution of x with respect to t. The derivative of x with respect to t is given by x' = e^(mt) (mC₁cos(βt) - βC₁sin(βt) + mC₂sin(βt) + βC₂cos(βt)).
Using the initial value condition x'(0) = -1, we substitute t = 0 and x' = -1 into the derivative equation. This gives us -1 = mC₁ - βC₁ + mC₂. Substituting the values of m, α, β, and C₁, we can solve for C₂ as follows:
-1 = (-2)(-0.5) - 2√3(-0.5) + mC₂
-1 = 1 + √3 + mC₂
-1 = 1 + √3 + (-2)(-0.5)C₂
-1 = 1 + √3 + C₂
Simplifying this equation, we find C₂ = -√3/3.
Finally, we have all the necessary values to write down the equation of motion with the initial value condition included. So, the equation of motion is x(t) = e^(-2t)(-0.5cos(2√3t) - (√3/3)sin(2√3t)).