How Is Damping Calculated in a Spring-Mass System?

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Homework Help Overview

The discussion revolves around calculating damping in a spring-mass system, specifically involving a mass of 0.0500 kg suspended from a spring with a spring constant of 25.0 N/m. The problem presents initial displacement and subsequent displacement after a certain time, with a damping force acting against the mass.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between forces acting on the mass, including spring force and damping force. Some express uncertainty about how to start solving the problem, while others suggest treating the governing equation as a differential equation. Various forms of motion equations are proposed, and there is discussion about the changing velocity and its implications for damping.

Discussion Status

The discussion is ongoing, with participants attempting to derive relationships and equations relevant to the problem. Some have suggested methods for solving the differential equation, while others are still grappling with the concepts and seeking clarification on their approaches.

Contextual Notes

Participants note a lack of prior instruction on damping, which contributes to their uncertainty. The original poster mentions the absence of a professor and expresses confusion about the problem setup and the calculations involved.

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Homework Statement



A egg with a mass 0.0500kg is suspended from a spring with a k = 25.0N/m... initially it starts off with a displacement of x = 0.300m but a force F = -bv acts against the egg and after t= 5.00s the displacement is x = 0.100m



Homework Equations



F = -bv
F = -kx
-kx = ma

w = [tex]\sqrt{}K/m[/tex]

ma = -kx - bv

The Attempt at a Solution



Well our prof didnt show up for class... but the homework is still due and i really have no idea were to start... i have never worked with damping... but i found out that

b = kg/s

Because F/v = kg/s

So if u calculate F at x = 0.300m you get 7.5N and then 2.5N for 0.1m...

thats a difference of 5... and that would be 1N/s loss when divided by time... which would make sense for the overall damping effect... (well I'm guessing... again never learned anything about damping) but if that is right... what 'v' would i use... because it is always changing as the spring looses energy...

So ya, I don't knoe if I'm even going in the right direction... but i am completely stumped on this one.
 
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miann said:

ma = -kx - bv


This is a differential equation. Treat it as such.
 
all I can think of is

x(t) = Acos(wt + [tex]\phi[/tex])

v(t) = -wAsin(wt + [tex]\phi[/tex])

a(t) = -w[tex]^{2}[/tex]Acos(wt + [tex]\phi[/tex])
 
Well... i can set it up as

-kx - b[tex]dx/dt[/tex] = m[tex]d^{2}x/dt^{2}[/tex]
 
but i can't think of the second equation so i can solve...
 
sorry i haven't done differential equations for about a year.. and even then they were easy ones... I'm trying to figure it out but i have no idea were to start.
 
Typically you want guess exponentials. x = Ae^(rt)
 
Try
http://www.krellinst.org/UCES/archive/modules/diffeq/node4.html
 
Last edited by a moderator:
kx = -bv

kx - -b[tex]\frac{dx}{dt}[/tex]

[tex]\int^{0.1}_{0.3}[/tex][tex]\frac{dx}{x}[/tex] = -[tex]\int^{t}_{0}[/tex][tex]\frac{k}{b}[/tex]dt


ln(0.1) - ln(0.3) = [tex]\frac{k}{b}[/tex](5.0s)

1.1 = [tex]\frac{(25.0N/m)}{b}[/tex](5.0s)

b = [tex]\frac{(25.0N/m)(5.0s)}{1.1}[/tex]

b = 113.64 N[tex]\bullet[/tex]s/m


is that right?
 
Last edited:

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