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Help with springs problem

by Hoppa
Tags: springs
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Hoppa
#1
May25-05, 12:10 PM
P: 38
hi i have this problem that i have to solve involving a spring and a mass attached to it and damping of the spring. i will first in this first post like put down the problem, and then in further posts i will put down how i am going with the problem, eg like what i have got etc, and um well any help along the way or if i get stuck be greatly appreciated thanks :)

problem:
A mass of 2kg is attached to one end of a spring, whose other end is fixed. the mass is free to move in one dimension, for which the spring constant is 18 Nm-1, and is unaffected by gravity. The motion of the system is damped by locating the mass within a viscous liquid causing a linear drag constant of 20 Nm-1s.

write down the position of the mass as a function of time after all the constants apart from constants of integration have been calculated.




thats the question, now in the next post, i will type out what i've got so far....
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Hoppa
#2
May25-05, 12:45 PM
P: 38
ok here the start of my working...


the equation of motion for the damped harmonic oscillator:
mx'' + bx' + kx = 0

are:
m = 2 kg
k = 18 Nm-1
b = 20 Nm-1
quasar987
#3
May25-05, 01:03 PM
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And I assume you don't know the general solution. To find it, notice that a function of the form

[tex]x(t) = Ce^{pt}[/tex]

where C, p are constants, is a solution. Find the value(s) of p by direct substitution into your diff. equ.

Yegor
#4
May25-05, 01:34 PM
P: 147
Help with springs problem

The general solution is [tex]x(t) =Ce^{-pt}[/tex]
Where C isn't constant, but it's like A*Sin[wt+fi]
quasar987
#5
May25-05, 01:41 PM
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Well that's what he was supposed to find out. You're just giving him the answer. The fact that the general solution is

[tex]x(t)=Ae^{-\gamma t}sin(\omega t + \phi)[/tex]

will flow from the fact that p is complex.
Yegor
#6
May25-05, 01:53 PM
P: 147
I'm really sorry. But if mx'' + bx' + kx = 0, and m = 2 kg; k = 18 Nm-1; b = 20 Nm-1, then the general solution isn't [tex]x(t)=Ae^{-\gamma t}sin(\omega t + \phi)[/tex]
, because friction is too "hard". I mean, that there must be aperiodic motion. Am i right?
Is [tex]x(t) = Ce^{pt}[/tex] right solution in all situations (when p is complex)??
quasar987
#7
May25-05, 02:14 PM
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Quote Quote by Yegor
I'm really sorry. But if mx'' + bx' + kx = 0, and m = 2 kg; k = 18 Nm-1; b = 20 Nm-1, then the general solution isn't [tex]x(t)=Ae^{-\gamma t}sin(\omega t + \phi)[/tex]
, because friction is too "hard". I mean, that there must be aperiodic motion. Am i right?
You're right. I hadn't bother to verify that k/m > (b/2m)² as I tought you had and that was basically what you were expressing in your last post when you said the gen. sol is

[tex]x(t)=Ce^{-pt}[/tex]

with C = A*sin(wt+phi)

Quote Quote by Yegor
Is [itex]x(t) = Ce^{pt}[/itex] right solution in all situations (when p is complex)??
p complex is only one possibility, arising when k/m > (b/2m)².

The form [itex]x(t) = Ce^{pt}[/itex] will indeed spawn the 3 different possibilities (damped, overdamped, underdamped) depending on the relation between k/m and (b/2m)².
Hoppa
#8
May27-05, 01:30 AM
P: 38
natural oscillation frequency of w = (square root) k/m
=18/2
=9 s-1
Hoppa
#9
May27-05, 01:38 AM
P: 38
damping = y = b/2m
= 20/2*2
= 20/4
= 5 s-1
Hoppa
#10
May29-05, 11:24 PM
P: 38
also is what i doing anywhere near correct?

any help would be appreciated.

thanks.



Mass of 2kg is attached to one end of a spring.
Spring constant = 18Nm-1
Damping, linear drag constant = 20Nm-1

Equation of motion for the damped harmonic oscillator:
mx’’ + bx’ + kx = 0

m = 2kg
k = 18 Nm-1
b = 20 Nm-1
.
w = Ö k
m
.
= Ö 18
2
= Ö 9
w = 3 s-1

Damping:
g = b
2m
= 20
2 x 2
g = 5 s-1

w2 < g2
Therefore the harmonic oscillator is overdamped since w2 < g2

x(t) = e-gt (Aeat + Be-at)

The position and velocity are:

x(t) = -5e-5t (Aeat + Be-at)

x’(t) = -5e-5t (aAeat - aBe-at)

For position as t = 0:
x(0) = e0 (Ae0 + Be0)
x(0) = 1 (A + B)
x(0) = A + B

For velocity as t = 0:
x’(0) = -5e0 (aAe0 - aBe-0)
x’(0) = -5 (aA - aB)
x’(0) = -5aA + 5aB

Substitute in initial conditions:
A + B = 0.5 (1)
-5aA + 5aB = 0 (2)
Rearrange equation (1):
A = 0.5 – B
Substitute into equation (2):
-5a(0.5 – B) + 5aB = 0
-2.5a + 5aB + 5aB = 0
10aB – 2.5a = 0
10aB = 2.5a
B = 2.5a
10a
B = 0.25
Substitute into equation (1)
A + 0.25 = 0.5
A = 0.25
Substitute into equation for position:

x(t) = e-5t (Aeat + Be-at)

x(t) = e-5t (0.25eat + 0.25e-at)

Velocity:

x’(t) = -5e-5t (0.25aeat – 0.25ae-at)

When velocity = 0

-5e-5t (0.25aeat – 0.25ae-at) = 0
Hoppa
#11
May29-05, 11:26 PM
P: 38
for the question im meant to have done:

1. Write down the position of the mass as a function of time after all the constants,
apart from constants of integration, have been calculated.
2. Obtain expressions for the constants of integration in terms of the initial conditions,
which are the position and velocity of the mass at time t = 0.
3. One initial condition is that, at t = 0, the mass is displaced by 0.5 m from the
equilibrium position. Evaluate the constants of integration and find the expression
for the position of the mass a function of time for each of the following
cases:
(a) The initial velocity is chosen to eliminate the slowly decaying exponential
from the solution. What is the value of this initial velocity?
(b) The initial velocity is zero.
Yegor
#12
May30-05, 10:14 AM
P: 147
i think it's all right with Your solution. Do You know how to express a in terms of w and g? I think it's also required. And i didn't find solution for 3 a).
OlderDan
#13
May30-05, 01:27 PM
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Quote Quote by Hoppa
for the question im meant to have done:

1. Write down the position of the mass as a function of time after all the constants,
apart from constants of integration, have been calculated.
2. Obtain expressions for the constants of integration in terms of the initial conditions,
which are the position and velocity of the mass at time t = 0.
3. One initial condition is that, at t = 0, the mass is displaced by 0.5 m from the
equilibrium position. Evaluate the constants of integration and find the expression
for the position of the mass a function of time for each of the following
cases:
(a) The initial velocity is chosen to eliminate the slowly decaying exponential
from the solution. What is the value of this initial velocity?
(b) The initial velocity is zero.
Question 3a suggests that your problem is for a driven harmonic oscillator, where the solution involves a decaying transient term, and a steady state term. If your problem is not a driven oscillator, there is no way to eliminate the decaying exponential except to eliminate the damping term. If it is driven, then an initial velocity can be chosen to eliminate the decaying transient term.
Yegor
#14
May30-05, 01:50 PM
P: 147
OlderDan, can You clarify it? When there is driven oscillator the solutuion consists of 2 terms. One is the solution of mx'' + bx' + kx = 0, and second is particular solution of driven oscillator equation (mx'' + bx' + kx = FSin[kt]). What does it mean "to eliminate the slowly decaying exponential from the solution"? Isn't it to find such A and B that solution of mx'' + bx' + kx = 0 equals zero? Then it could be made also for damped oscillator. Where am i wrong?
OlderDan
#15
May30-05, 11:08 PM
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P: 3,031
Quote Quote by Yegor
OlderDan, can You clarify it? When there is driven oscillator the solutuion consists of 2 terms. One is the solution of mx'' + bx' + kx = 0, and second is particular solution of driven oscillator equation (mx'' + bx' + kx = FSin[kt]). What does it mean "to eliminate the slowly decaying exponential from the solution"? Isn't it to find such A and B that solution of mx'' + bx' + kx = 0 equals zero? Then it could be made also for damped oscillator. Where am i wrong?
The transient solution to the driven oscillator includes the solution to the homogeneous equation, but it is not just the homogeneous solution. It is also affected by the driving force. I know this is not exactly the situation in this problem, but if you look at this link that has the solution for the underdamped case

http://hyperphysics.phy-astr.gsu.edu.../oscdr.html#c1

you can see that the transient term has factors involving the magnitude, frequency, and phase of the driving force.

Without working out the complete solution, I'm not sure exactly what the transient term will look like for the overdamped case, but there will still be factors in the amplitude of the transient term involving the driving force. It is the mixing of the driving effects and the homogeneous contribution that makes it possible to set the transient amplitude to zero. If all you have is the undriven damped oscillator, there will always be a real part in the exponential factor equal to -b/2m (b is c in the referenced equations). The only way to eliminate the decaying exponential is to let b go to zero.


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