Mass-spring-damper system solve for x(t) using power series

[maciejj]

Homework Statement

A mass of 10kg is suspended from vertical spring of stiffens 100N/m and is provided with dashpot damper having damping coefficient of 1000Ns/m.
The mass is pulled down the distance of 4cm from its equilibrium position and than released.
Establish the differential equation of motion and solve using power series for variation in displacement with time.

Homework Equations

The Attempt at a Solution

Started with
mx''(t)=-kx(t)-cx'(t)
and initial conditions
t0=0,x(t0)=4cm=0.04m, x'(t0)=0,x''(t0)=0 here assume that velocity and acceleration=0
at t=0
after solving using power series got:
x(t)=a0+a1x+a2x2+a3x3+... a0=0.04 but all the rest of coefficients =0

That can be right ,please help

 

[NeroKid]

its not 0 for the rest for coefficient , u just have to treat the equation as some function and take its derivative, and since u know x' and x u know x'' by the equation , taking the derivative will get u the equation consist of x''' , x'' and x' and x'', x' have already been known

 

[maciejj]

mx^''=-kx-cx'
x^''+c/m x^'+k/m x=0
Putting known values into equation:
x^''+100/10 x^'+1000/10 x=0
x^''+10x^'+100x=0
Now solving using power series:
Let assume that:
x(t)=a_0+a_1 t+a_2 t^2+a_3 t^3+..
t_0=0,x_0=0.04 so a_0=0.04 From initial conditions

x^' (t)=a_1+〖2a〗_2 t+〖3a〗_3 t^2+〖4a〗_4 t^3+..
t_0=0,〖x'〗_0=0 so a_1=0 From initial conditions

x^'' (t)=〖2a〗_2+〖6a〗_3 t+〖12a〗_4 t^2+20a_5 t^3+..
t_0=0,〖x''〗_0=0 so a_2=0 From initial conditions
Putting these into differential equation
x^''+10x^'+100x=0
〖(2a〗_2+〖6a〗_3 t+〖12a〗_4 t^2+20a_5 t^3+..)+10(a_1+〖2a〗_2 t+〖3a〗_3 t^2+〖4a〗_4 t^3+..)+
+100(a_0+a_1 t+a_2 t^2+a_3 t^3+..)=0

Rearranging
100a_0+10a_1+2a_2+(100a_1+20a_2+6a_3 )t+(100a_2+30a_3+12a_4 ) t^2+
+(100a_3+40a_4+20a_5 ) t^3=0
And a_0=0.04 ,a_1=0,a_2=0 -From initial conditions
0.04+(100a_1+20a_2+6a_3 )t+(100a_2+30a_3+12a_4 ) t^2+(100a_3+40a_4+20a_5 ) t^3=0

this is where I m now
but don't konw how to get rest of coefficents

 

[LCKurtz]

Are you trying to solve this by power series for some reason? Although that should work, it is certainly the hard way to do a constant coefficient DE.

 

[maciejj]

hi ,yes its one of task for my math assignment and its specified to solve it using power series

 

[NeroKid]

actually x''(0) is not zero since the object is at the amplitude it should have maximum accelaration toward equilibrium point

 

[Ray Vickson]

Homework Statement

A mass of 10kg is suspended from vertical spring of stiffens 100N/m and is provided with dashpot damper having damping coefficient of 1000Ns/m.
The mass is pulled down the distance of 4cm from its equilibrium position and than released.
Establish the differential equation of motion and solve using power series for variation in displacement with time.

Homework Equations

The Attempt at a Solution

Started with
mx''(t)=-kx(t)-cx'(t)
and initial conditions
t0=0,x(t0)=4cm=0.04m, x'(t0)=0,x''(t0)=0 here assume that velocity and acceleration=0
at t=0
after solving using power series got:
x(t)=a0+a1x+a2x2+a3x3+... a0=0.04 but all the rest of coefficients =0

That can be right ,please help

Show your work. Without some indication of where you went wrong, it would be impossible to assist you.

RGV

 

[maciejj]

Hi ,have uploaded a pdf file showing what I have done.
So what initial conditions would you use?

 

[Ray Vickson]

Hi ,have uploaded a pdf file showing what I have done.
So what initial conditions would you use?

People keep telling you that x''(0) is nonzero, but you don't listen. The DE x'' = -10 x' - 100x gives x''(0) = -100*4/100 = -4.

That will give you an equation of the form C1*t + C2*t^2 + C3*t^3 + ... = 0, where the C1, C2, C3,... are linear combinations of your a1, a2, a3, ... . The equation is supposed to be an identity in t, so that means that all the coefficients must vanish; that is, we must have C1 = 0, C2 = 0, C3 = 0, ... . Solving those will give you the coefficients a1, a2, a3, ... in the expansion of x(t).

RGV

 

[maciejj]

have changed x''(0)
is that correct now?