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Damped Oscillator help please!

  • Thread starter Full Beige
  • Start date
  • #1

Homework Statement


Hi guys the question is: a mass spring-damper system is positioned between two rigid surfaces, if mass m = 200g, spring constant k = 80 Nm-1, and damping pot of coefficient 65 gs-1. The mass is pulled 5cm down from its equilibrium position and then released. What is the period of motion assuming the system is conservative?


Homework Equations


period T = 2pi / omega >> (eq1)

and angular frequency omega^2 = (k / m) - (b/2m)^2 >>(eq2)

where k = spring constant
m = mass
b = coefficient of damping


The Attempt at a Solution



using eq2 omega^2 = 80/200 - (65 / 2x 200)^2 => 0.4 -0.026 = 0.374

omega = sqrroot 0.374 = 0.061

plug this into eq1 => T = 2pi / 0.061 = 10.30 s

this value seems to high for me, can anybody see what I've done wrong? thanks in advanced.

Beige
 
Last edited:

Answers and Replies

  • #2
175
1
It looks like you need to be a little more careful with your units...

-Kerry
 
  • #3
thanks I saw the mistake, I didnt convert the g to kg. cheers for your hint anyhows thanks

beige
 
  • #4
Hi guys Im back again, I'm kind of stuck on part c of this question:

how long does it take for the inital mechanical energy of the damped system to halve its value?

part b asked how long it would take for the amplitude to halve, which i worked out to be 4.27s.

I have used 1/2 x m x Vm^2 >> where Vm = omega x Xm (Xm = amplitude) this gives me >> (0.5 x 0.2 (4.27 x 19.19)^2) / 2 = 364.29 J

then I used T = 2pi x sqroot(m/c) for some reason I put c = 364.29 and hence got T = 3.44s.

I'm kind of lost if I'm anywhere near the answer. any help would be greatly appreciated.

thanks

beige
 
  • #5
175
1
Have you determined how much energy is in the system at the beginning? Determine the amount of energy in the system just before it is released. What equations do you need to use for that? Also, I don't think your Vm calculation is valid (try using the energy approach).

-Kerry

p.s. LaTeX works on these forums, and it helps to make the math a lot easier to read (click on the [tex]\Sigma[/tex] along the top of the text box where you post)
 
  • #6
o.k looking at my notes i can see that Mechanical energy E{t} = [tex]\frac{1}{2}[/tex]kxm2 e-bt/m so using this formula I get: [tex]\frac{1}{2}[/tex]E{t}= [tex]\frac{1}{2}[/tex]kxm2 e-bt/m plugging in the values:

[tex]\frac{1}{2}[/tex]E{t}= [tex]\frac{1}{2}[/tex]80x4.27m2 e-0.065x0.31/0.2 = 659.30 / 2 => 329.65 J

and then rearranging the same formula for t:

t = -[tex]\frac{m}{b}[/tex] ln [tex]\frac{2e}{kxm^2}[/tex] I get = > -3.07 ln 0.45 = 2.45s

thanks for laTex tip and thanks for the help Kloux. Does this look any better?
 
  • #7
175
1
Can you explain the variables and the values you choose for them in this equation? I'm particularly interested in the t in the exponential term. It appears that you chose 0.31 for this value. Why?

I think you can find your solution using the equation from your last post, but not using that approach. Understanding what t is will be a big help! From the second half of your last post, it seems that you do understand it, but maybe you were confused when you were solving for the energy in the system just before it is released.

Think about what the notation E(t) means. I think you're on the right track.

Here's another hint: The answer will depend only on b and m.

-Kerry
 
  • #8
m = mass, b = damping constant, t = period ( not clear from my notes so I assumed was period), k = spring constant, xm2 = oscillation amplitude.

period t I used the forumla t = 2pi[tex]\sqrt{m/k}[/tex] which gave me => 0.31s

for xm2 = oscillation amplitude I used t = ((-2 (m) ln 1/2) / b) = 4.27s.

does that make it any clearer? and thanks for taking a look Kerry

beige
 

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