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 and the right hand support oscillates at where is the angular frequency, and is the separation between the supports when the system is at rest.

If is the position of the mass relative to the origin, find a ODE for 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

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 but I have no idea when it's placed in between.

How do I start? Just a little nudge in the right direction please.

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!

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 ... I don't know what it should equal.

Quote:

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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 ... I don't know what it should equal.

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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

Quote:

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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!

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

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