Can an Object Attached to a Spring Return to its Equilibrium Position Only Once?

[OptimusPrime]

An object weighing 34 lbs. is attached to a spring with spring constant 1 lb/ft. This spring-mass system is subject to a damping forece numerically equal to twice the velocity. At time t=0 the object is lowered 3 inches below its equilibrium position and given an initial velocity of 1 ft./sec in the upward direction.

Show that the object will pass through it equilibrium position exactly once and then creep back towards its equilibrium position.

Thank you


A Response by Kevin:
2004-04-21 at 13:41GMT

This is what I have so far. Please help

So the first thing we need to do is formulate the differential equation. Based on one of Newton's Laws of Motion (forgot which one offhand), the sum of the forces on a body equals the product of the mass and acceleration of the body. If we denote the displacement (in feet) of the mass from its equilibrium position by x (with positive x corresponding to the mass lying below its equilibrium position), then

spring force = -k*x, k = 1 lb/ft (Hook's Law)
damping force = -c*(dx/dt), c = 2 lb/(ft/s)

Since acceleration is the second time derivative of displacement, we arrive at

m*(d^2x/dt^2) = -k*x - c*(dx/dt)

Letting ' denote differentiation by t, we obtain the differential equation

m*x'' + c*x' + k*x = 0

with initial conditions

x(0) = x0 = 3 inches = 1/4 ft
x'(0) = v0 = -1 ft/s

 

[arildno]

Try an exponential solution, e^(i*at), and see what get.. (i sqrt(-1))

 

[OptimusPrime]

thanks

Thanks for the response,

I've been at this problem for hours. I don't know how to apply what you said into the problem. Please tell me how.

 

[arildno]

1. of all it's Newton's second law.
2. You've got what's called a linear differential equation with constant coeffecients (LDEC)
3. It has been shown, that every (LDEC) has exponential solutions, if you include the complex exponential in your definition of the exponential function.
4. Perhaps you haven't heard of the c.ex. before?
In that case:
e^(i*w)=cos(w)+i*sin(w), i=sqrt(-1)
5. If you haven't seen complex numbers before, don't be scared!
Think of a plane with the real number line as the x-axis, the y-axis is the imaginary line.
Any point in the (x,y) plane is called a complex number, and is written as x+i*y
All operations of addition and multiplication is as before, just remember that i*i=-1

TO THE PROBLEM:
6. Differentiation is just the same with a complex function, than with a real function.

7. We have, with the trial solution y=e^(i*at):
y'=i*a*e^(i*at), y''=(i*a)^(2)*e^(i*at)=-a^(2)*e^(i*at)

8. We enter these relations into the diff. eq. and get:
(-m*a^(2)+i*c*a+k)*e^(i*at)=0

9. If our trial function shall be a solution, it has to be a solution at each value of t.
Therefore, the bracketed coefficient must be 0, and we have the following equation:
-m*a^(2)+i*c*a+k=0
10. This is a second degree polynomial in a; we can now find out specifically which
values of a is allowed, given k,m,c.
11.
Using the quadratic formula, we get:
a=(-i*c+-sqrt(-c^(2)+4mk))/(-2m):
a1=i*c/m+sqrt(4mk-c^(2)))/(2m)
a2=i*c/m-sqrt(4mk-c^(2)))/(2m)

Give a message if it's something you're wondering about, or want to proceed..

 

[OptimusPrime]

Thanks so much,

Yes I want to proceed. I'm not sure how much of the problem is left to complete?

 

[arildno]

Sorry for not giving a reply sooner.
There's a bit more, but basically, we are finished.
1. If you evaluate the square root, you''ll see that it is greater than 0

2.Let's put in a1 in our trial solution:
e^(i*a1t)=e^(-c/m*t)*(cos(sq.*t)+i*sin(sq.*t)), sq.=sqrt(4mk-c^(2))/2m
If you use a2, the only difference will be a minus sign in front of the i*sin(sq.*t) term.

3. To find two real, independent, solutions, use the "real" part and the "imaginary" part.
The real part is: R(t)=e^(-c/m*t)*cos(sq.*t)
The imaginary part is: I(t)=e^(-c/m*t)*sin(sq.*t)

4. You can now write any solution as u*R(t)+v*I(t), where u and v is determined by the initial conditions.