How Do You Calculate the Damping Coefficient in a Mass-Spring System?

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SUMMARY

The discussion focuses on calculating the damping coefficient (b) in a mass-spring system with a mass of 0.2 kg and a spring constant of 80 N/m. The damped frequency is specified as √(3)/2 of the undamped frequency. Participants clarify the distinction between natural frequency (ωn = √(k/m)) and damped frequency (ωd), emphasizing the correct use of subscripts in equations. The final approach involves rewriting the equation for displacement as x(t) = Ae^(γt)cos(ωdt) and solving for b using the relationship between ω and γ.

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  • Understanding of mass-spring systems and their dynamics
  • Familiarity with differential equations
  • Knowledge of natural frequency and damped frequency concepts
  • Ability to manipulate equations involving exponential and trigonometric functions
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  • Study the derivation of the damping coefficient in mass-spring systems
  • Learn about the differences between natural frequency (ωn) and damped frequency (ωd)
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AsadaShino92
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Homework Statement


An object of mass 0.2kg is hung from a spring whose spring constant is 80N/m. The object is subject to a resistive force given by -bv, where v is it's velocity in meters per second.

If the damped frequency is √(3)/2 of the undamped frequency, what is the value of b?

Homework Equations


F=ma
ω=√k/m

The Attempt at a Solution


I tried to write the sum of the forces of the system and got ∑F=-kx-bv=ma
I rewrote it as -kx=b(dx/dt)+m(d^2x/dt^2)

Now I don't have much experience with differential equations but I know the solution is x(t)=Ae^(γt)cos(ωt) where γ=(-b/2m). I also know that the damped frequency is (√(3)/2)√k/m given from the problem. I not sure where to go from here. I am supposed to use the solution and solve for b? Any help would be appreciated.
 
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AsadaShino92 said:
Now I don't have much experience with differential equations but I know the solution is x(t)=Ae^(γt)cos(ωt) where γ=(-b/2m).
What is the value of ω in terms of k, m and/or b? You know γ(b,k,m) but you don't know ω(b,k,m)?
 
rude man said:
What is the value of ω in terms of k, m and/or b? You know γ(b,k,m) but you don't know ω(b,k,m)?

So ω in terms of k,m, and b would be √((ω^2)-(γ^2)) right?
 
Going back to your 1st post, you are mixing up two ω's. One is the natural frequency ωn = √(k/m). the other is the damped frequency which is ωd. The ω in your cos argument should be the latter. The idea is that ωd < ωn as your problem statement gives. Don't use ω again, use the two above.
Having said that, to answer your question
So ω in terms of k,m, and b would be √((ω^2)-(γ^2)) right?
is correct IF you use the right omegas. You can't say x = x + a, a ≠ 0, can you?
 
Last edited:
rude man said:
Going back to your 1st post, you are mixing up two ω's. One is the natural frequency ωn = √(k/m). the other is the damped frequency which is ωd. The ω in your cos argument should be the latter. The idea is that ωd < ωn as your problem statement gives. Don't use ω again, use the two above.
Having said that, to answer your question is correct IF you use the right omegas. You can't say x = x + a, a ≠ 0, can you?

Yeah I agree with that. I seem to be getting the 2 omega's confused with each other. So my solution should actually be expressed as x(t)=Ae^(γt)cos(ωdt), where ωd refers to the damped frequency. Sorry for making it look messy but I wasn't sure how to write the subscript d. Plugging in √((ω^2)-(γ^2)) for ω damping I can solve for my value of b. Is that the right idea?
 
AsadaShino92 said:
Yeah I agree with that. I seem to be getting the 2 omega's confused with each other. So my solution should actually be expressed as x(t)=Ae^(γt)cos(ωdt), where ωd refers to the damped frequency. Sorry for making it look messy but I wasn't sure how to write the subscript d. Plugging in √((ω^2)-(γ^2)) for ω damping I can solve for my value of b. Is that the right idea?
1. You're still using ω instead of wd or ωn. Don't.
2 . Picking the correct omegas, rewrite your equation; this time make it a real equation with an = sign and everything.
3. It's easy to make subscripts or superscripts. See the "x2" and the "x2" on the toolbar where you got your ω?
 
Sorry for the late reply. Yes I found it, thanks for pointing that out.
rude man said:
3. It's easy to make subscripts or superscripts. See the "x2" and the "x2" on the toolbar where you got your ω?
 

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