Spring damper system equation of motion

[btnsteve]

Homework Statement

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.

Homework Equations

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}
m_eq = m₂ + \frac{I}{r^{2}} where m₂is 4*m₁

I then substituted the numbers in and found M_eq= 190

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

C_c = 2*m*W_n
W_n = \sqrt{\frac{K}{M}}
W_n= \sqrt{\frac{20000}{190}} W_n = 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.

 

[Simon Bridge]

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

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.

Could the Q factor help you there?

 

[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

 

[Simon Bridge]

Oh I see - you shifted to an equivalent system with a different mass to keep the torques the same.

Q factor ... and more on the properties of the damped harmonic oscillator.
http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

Also look at "step response".

 

[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)

 

[Simon Bridge]

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

 

[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.

 

[Simon Bridge]

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,

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

so energy is somewhat proportional to velocity squared so if I could find velocity I could find the time through equation of motion.

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?