Harmonic motion with damping

  • #1

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.
 

Answers and Replies

  • #2
rude man
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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)?
 
  • #3
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?
 
  • #4
rude man
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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?
 
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  • #5
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?
 
  • #6
rude man
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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?
 
  • #7
Sorry for the late reply. Yes I found it, thanks for pointing that out.
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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