# Spring damper system equation of motion

1. Nov 16, 2013

### btnsteve

1. The problem statement, all variables and given/known data
A 10 kg block is displaced 20 mm and released. If damping coefficient is 100 N.s/m,
how many cycles will be executed before amplitude is reduced to 1 mm or below? The stiffness
of the spring is k=20000 N/m.

2. Relevant equations

3. The attempt at a solution
I first moved the mass to the inner radius and equated the Kinetic Energy of the system.
Ke1 = Ke2
Where i found m2 is 4m1

Next i equated the kinetic energy of the system and equated that to :
$\frac{1}{2}$*m$_{eq}$*v$^{2}$
$\frac{1}{2}$*m$_{2}$v$^{2}$ + $\frac{1}{2}$I$\frac{V^{2}}{r^{2}}$= $\frac{1}{2}$* m$_{eq}$*v$^{2}$
meq = m2 + $\frac{I}{r^{2}}$ where m2is 4*m1

I then substituted the numbers in and found Meq= 190

Next to find the amplitude i found the damping ratio of the system
$\zeta$ =$\frac{c}{Cc}$

Cc = 2*m*Wn
Wn = $\sqrt{\frac{K}{M}}$
Wn= $\sqrt{\frac{20000}{190}}$ Wn = 10.26
Cc = 2(190) * (10.26) = 3899 ∴ $\zeta$ = $\frac{100}{3899}$
$\zeta$ = 0.0256
Under damped system E.O.M =
X(t) = e$^{-\zeta*W_{n}*t}$ { x$_{o}$Cos(w$_{d}$t) + $\frac{x^{.}+W_{n}*X_{0}}{w_{d}}$*Sin(w$_{d}$t) }
I'm trying to find the t value that would make X(t) be less than 1mm, i'm not sure how i would do that without just picking random values of t, as the equation doesn't seem solvable just for t.

Last edited: Nov 16, 2013
2. Nov 16, 2013

### Simon Bridge

Why did you move the mass to the inner radius and what is m2?
Anyway:

3. Nov 16, 2013

### btnsteve

I moved the mass to the inner radius i guess to simplify the system, honestly not sure, its the way I've been taught this semester to do it.

So if i had the original system the mass would be M1
But now I've moved it to the inner radius The mass is now M2
For it to still be the same $\frac{1}{2}$M1*V1$^{2}$ = $\frac{1}{2}$M2*V2$^{2}$

Where V1 = $\dot{\theta}$2r
Where V2 = $\dot{\theta}$r
Therefore
$\frac{1}{2}$M1*($\dot{\theta}$2r$)^{2}$ = $\frac{1}{2}$M2*($\dot{\theta}$r)$^{2}$
Through Cancelling 4M1 = M2

I'm not sure what the Q factor is

4. Nov 16, 2013

### Simon Bridge

5. Nov 17, 2013

### btnsteve

Still not 100% sure on this Q factor, hard to see how it applies without an example.

Q = $\frac{1}{2\zeta}$
Q = $\frac{1}{2*0.0256}$ Q = 19.53

Q = 2$\pi$*$\frac{Energy Stored}{Energy Lost Per Cycle}$

$\frac{Q}{2\pi}$ = $\frac{Energy Stored}{Energy Lost Per Cycle}$

3.11 = $\frac{Energy Stored}{Energy Lost Per Cycle}$

Can you give a hint at what the next step would be, would i find the energy at the start? By using
T(t) + V(t) = E, Where T is the kinetic energy and V is the potential, i guess at t = 0 there is no kinetic energy.

Therefore $\frac{1}{2}$kx$_{0}$$^{2}$ = E(0)

6. Nov 17, 2013

### Simon Bridge

How does the energy relate to the amplitude of the oscillations?

7. Nov 18, 2013

### steve2510

Well as the energy is lost the amplitude decreases starting off at a maximum I guess, so there must be a time where the amount of energy lost means the amplitude falls below 2mm, i mean if I multiply both sides by 3.18 I get that 3.18 x energy lost per cycle = energy stored, at first I thought that meant it oscillates for 3.18 cycles but thst seems too little when the damping is that small. I guess directly the potential energy at the start is converted into kinetic energy so energy is somewhat proportional to velocity squared so if I could find velocity I could find the time through equation of motion.

8. Nov 18, 2013

### Simon Bridge

Which means that you need to know how the amplitude related to the energy ... i.e. is it an inverse-square law?

You can do it more directly using potential energy ... all the kinetic energy ends up stored in the spring, and the energy stored in a spring is related to how far it is compressed ... which is what you want to know about.

When the amplitude is 2mm, then the system has lost a certain percentage of amplitude ... which relates to a certain percentage of energy, and you have an equation for the rate of energy loss with time.

You should check the extent of the damping though - is this underdamped, critically damped, what? And what does that mean for the general motion?