- #1
CraigH
- 222
- 1
What is the general solution to the differential equation describing a mass-spring-damper?
t=time
x= extension of spring
M=Mass
K=Spring Constant
C=Damping Constant
g= acceleration due to gravity
Spring has 0 length under 0 tension
Spring has 0 extension at t = 0
If the Force downwards due to the mass equals:
Force downwards = M*g
And this is opposed by tension in the spring:
Force upwards = - K*x
And there is a 3rd force acting on the system caused by the damper. This allways acts in the opposite direction as velocity.
Force = - C*(dx/dt) //because damping is proportional to velocity.
So the overall force acting on the mass equals:
Net Force = M*g - K*x - C*(dx/dt)
This net force causes an acceleration in the direction of the force:
F=Ma so...
M*(d2x/dt2) = M*g - K*x - C*(dx/dt)
\\you can see that this will oscillate, because when x is 0 the acceleration will be positive, when M*g= K*x + C*(dx/dt), the acceleration will be 0, and then become negative, and then the velocity will become 0, and then negative, and then the x will become 0 again, and it will repeat.
So the differential equation describing this system is:
M*(d2x/dt2) + C*(dx/dt) + K*x = Mg
Mx''+Cx'+Kx=Mg
This is a second order linear non homogeneous differential equation.
Mx''+Cx'+Kx=g(t)
where g(t) is a constant.
Now my question is, how do I solve it for x?
I tried using the undetermined coefficients method, but I do not get an equation that implies that x is oscillating. Shouldn't I get a sin or cos somewhere in the answer?
Here is what I tried:
corresponding Homogeneous Equation:
Mx''+Cx'+Kx=0
characteristic equation = M*(r^2) + C*r + K
General solution to the homogeneous equation:
C1(e^pt)+C2(e^qt)
where
p=(-C+SQRT((C^2)-4*M*K))/2M
q=(-C-SQRT((C^2)-4*M*K))/2M //from the quadratic equation
Now to calculate the particular solution:
Mx''+Cx'+Kx=Mg
The right hand side is a polynomial of degree 0. try a polynomial as the solution
x=Ax^2 + Bx + C
x'=2Ax +B
x''=2A
Sub these into the differential equation and you end up with:
x=Mg/K
So the general solution to the differential equation is:
x(t) = C1(e^pt)+C2(e^qt) + Mg/K
This can't be correct, because x should oscillate over time.
C1, C2, p, q, ang Mg/K are all constants so this equation does not describe an oscillation.
Where did I go wrong?
Thanks!
t=time
x= extension of spring
M=Mass
K=Spring Constant
C=Damping Constant
g= acceleration due to gravity
Spring has 0 length under 0 tension
Spring has 0 extension at t = 0
If the Force downwards due to the mass equals:
Force downwards = M*g
And this is opposed by tension in the spring:
Force upwards = - K*x
And there is a 3rd force acting on the system caused by the damper. This allways acts in the opposite direction as velocity.
Force = - C*(dx/dt) //because damping is proportional to velocity.
So the overall force acting on the mass equals:
Net Force = M*g - K*x - C*(dx/dt)
This net force causes an acceleration in the direction of the force:
F=Ma so...
M*(d2x/dt2) = M*g - K*x - C*(dx/dt)
\\you can see that this will oscillate, because when x is 0 the acceleration will be positive, when M*g= K*x + C*(dx/dt), the acceleration will be 0, and then become negative, and then the velocity will become 0, and then negative, and then the x will become 0 again, and it will repeat.
So the differential equation describing this system is:
M*(d2x/dt2) + C*(dx/dt) + K*x = Mg
Mx''+Cx'+Kx=Mg
This is a second order linear non homogeneous differential equation.
Mx''+Cx'+Kx=g(t)
where g(t) is a constant.
Now my question is, how do I solve it for x?
I tried using the undetermined coefficients method, but I do not get an equation that implies that x is oscillating. Shouldn't I get a sin or cos somewhere in the answer?
Here is what I tried:
corresponding Homogeneous Equation:
Mx''+Cx'+Kx=0
characteristic equation = M*(r^2) + C*r + K
General solution to the homogeneous equation:
C1(e^pt)+C2(e^qt)
where
p=(-C+SQRT((C^2)-4*M*K))/2M
q=(-C-SQRT((C^2)-4*M*K))/2M //from the quadratic equation
Now to calculate the particular solution:
Mx''+Cx'+Kx=Mg
The right hand side is a polynomial of degree 0. try a polynomial as the solution
x=Ax^2 + Bx + C
x'=2Ax +B
x''=2A
Sub these into the differential equation and you end up with:
x=Mg/K
So the general solution to the differential equation is:
x(t) = C1(e^pt)+C2(e^qt) + Mg/K
This can't be correct, because x should oscillate over time.
C1, C2, p, q, ang Mg/K are all constants so this equation does not describe an oscillation.
Where did I go wrong?
Thanks!