# Double Spring Mass System

• Apr 11th 2011, 12:48 PM
ShazamV4
Double Spring Mass System
A mass (M) is connected to two oscillating supports with springs (mass in the middle, then single spring on each side connecting to a moving support). The left support oscillates at \$\displaystyle X_1=Acos(wt)\$ and the right hand support oscillates at \$\displaystyle L_0+Bwsin(wt)\$ where \$\displaystyle w\$ is the angular frequency, and \$\displaystyle L_0\$ is the separation between the supports when the system is at rest.

If \$\displaystyle X(t)\$ is the position of the mass relative to the origin, find a ODE for \$\displaystyle X(t)\$ and find the location of the mass when at equillibrium/rest. That is, find the position of the mass when its velocity and acceleration are both 0, assuming that A=0 and B=0. Both springs satisfy Hooke's law with constants of \$\displaystyle k_1 and k_2\$

Thats the problem statement.

I know that this has to be a second order ODE because it's oscillating. Also, I know the form of a mass on the end of a single spring is \$\displaystyle mx''(t)+kx(t)=0\$ but I have no idea when it's placed in between.

How do I start? Just a little nudge in the right direction please.
• Apr 11th 2011, 01:56 PM
derdack
This is a forced vibration problem with one degree of freedom.

x''+(k/m)x=(Acos(ot)+L0+Bsin(ot))/m

xp=Ccos(ot)+Dsin(ot), xp'=-C*o*sin(ot)+D*o*cos(ot), xp''=-C*o^2*cos(ot)-D*o^2*sin(ot).

k/m=o1^2 angular frequency.

Substitution of xp and xp'' into diff equation you can find constants and this can determine xp.

Using initial conditions - velocity and acceleration are both 0, you can find xh now. xh=A1*cos(o1*t)+B1*sin(o1*t), xh'=...,Substituting t=0 you can find xh

x=xh+xp

Kind regards!
• Apr 13th 2011, 03:41 AM
ShazamV4
Awesome... but we were just told that he forgot to tell us that there is a friction which is proportional to the velocity, where b is for friction. So that means it would be like \$\displaystyle mx''+bx'+kx=\$... I don't know what it should equal.
• Apr 13th 2011, 04:45 AM
topsquark
Quote:

Originally Posted by ShazamV4
Awesome... but we were just told that he forgot to tell us that there is a friction which is proportional to the velocity, where b is for friction. So that means it would be like \$\displaystyle mx''+bx'+kx=\$... I don't know what it should equal.

The RHS will be the same as it was in derdack's post. I would note, though, that the standard way of writing the friction term would be to add a -bx' as the friction is opposing the direction of motion.

-Dan
• Apr 13th 2011, 11:07 AM
ShazamV4
Quote:

Originally Posted by derdack
This is a forced vibration problem with one degree of freedom.

x''+(k/m)x=(Acos(ot)+L0+Bsin(ot))/m

xp=Ccos(ot)+Dsin(ot), xp'=-C*o*sin(ot)+D*o*cos(ot), xp''=-C*o^2*cos(ot)-D*o^2*sin(ot).

k/m=o1^2 angular frequency.

Substitution of xp and xp'' into diff equation you can find constants and this can determine xp.

Using initial conditions - velocity and acceleration are both 0, you can find xh now. xh=A1*cos(o1*t)+B1*sin(o1*t), xh'=...,Substituting t=0 you can find xh

x=xh+xp

Kind regards!

What was the benefit of dividing by \$\displaystyle m\$? Can't I just leave \$\displaystyle m\$ in there, get solve for \$\displaystyle x_h(t)\$ (the homogeneous portion of the ODE), then solve for \$\displaystyle x_p(t)\$... or is there a big shortcut/trick that I'm missing here.

Also, how do you solve for o, in xp(t)