Register to reply 
Help with springs problemby Hoppa
Tags: springs 
Share this thread: 
#1
May2505, 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 Nm1, 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 Nm1s. 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.... 


#2
May2505, 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 Nm1 b = 20 Nm1 


#3
May2505, 01:03 PM

Sci Advisor
HW Helper
PF Gold
P: 4,771

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. 


#4
May2505, 01:34 PM

P: 149

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] 


#5
May2505, 01:41 PM

Sci Advisor
HW Helper
PF Gold
P: 4,771

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. 


#6
May2505, 01:53 PM

P: 149

I'm really sorry. But if mx'' + bx' + kx = 0, and m = 2 kg; k = 18 Nm1; b = 20 Nm1, 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)?? 


#7
May2505, 02:14 PM

Sci Advisor
HW Helper
PF Gold
P: 4,771

[tex]x(t)=Ce^{pt}[/tex] with C = A*sin(wt+phi) 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. 


#8
May2705, 01:30 AM

P: 38

natural oscillation frequency of w = (square root) k/m
=18/2 =9 s1 


#9
May2705, 01:38 AM

P: 38

damping = y = b/2m
= 20/2*2 = 20/4 = 5 s1 


#10
May2905, 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 = 18Nm1 Damping, linear drag constant = 20Nm1 Equation of motion for the damped harmonic oscillator: mx + bx + kx = 0 m = 2kg k = 18 Nm1 b = 20 Nm1 . w = Ö k m . = Ö 18 2 = Ö 9 w = 3 s1 Damping: g = b 2m = 20 2 x 2 g = 5 s1 w2 < g2 Therefore the harmonic oscillator is overdamped since w2 < g2 x(t) = egt (Aeat + Beat) The position and velocity are: x(t) = 5e5t (Aeat + Beat) x(t) = 5e5t (aAeat  aBeat) 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  aBe0) 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) = e5t (Aeat + Beat) x(t) = e5t (0.25eat + 0.25eat) Velocity: x(t) = 5e5t (0.25aeat 0.25aeat) When velocity = 0 5e5t (0.25aeat 0.25aeat) = 0 


#11
May2905, 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. 


#12
May3005, 10:14 AM

P: 149

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).



#13
May3005, 01:27 PM

Sci Advisor
HW Helper
P: 3,033




#14
May3005, 01:50 PM

P: 149

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?



#15
May3005, 11:08 PM

Sci Advisor
HW Helper
P: 3,033

http://hyperphysics.phyastr.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. 


Register to reply 
Related Discussions  
Springs in series and parallel : Problem for U  Introductory Physics Homework  4  
Physics problem about springs.. PLEASE HELP!  Advanced Physics Homework  2  
Conceptual Problem with Springs and Energy  Introductory Physics Homework  4  
Physic problem... springs  Introductory Physics Homework  1  
Mass and springs problem ASAP  Introductory Physics Homework  3 