A mass m=4 is attached to both a spring with spring constant k=101 and a dash-po
ID: 2005572 • Letter: A
Question
A mass m=4 is attached to both a spring with spring constant k=101 and a dash-pot with damping constant c=4.
The ball is started in motion with initial position x0=2 and initial velocity v0=8 .
Determine the position function x(t) .
x(t)=?
Note that, in this problem, the motion of the spring is underdamped, therefore the solution can be written in the form x(t)=c1e^(-t)cos(1t-1) . Determine c1, 1, and 1.
c1=?
1=?
=?
1=?
Graph the function x(t) together with the "amplitude envelope" curves x=-c1e^(-t) and x=c1e^(-t).
Now assume the mass is set in motion with the same initial position and velocity, but with the dashpot disconnected ( so c=0). Solve the resulting differential equation to find the position function u(t).
In this case the position function u(t) can be written as u(t)=c0cos(0t-0). Determine c0, 0,and 0.
c0=?
0=?
0=?
Explanation / Answer
Part 1:
set up the IVP in the following manner: mu''+cu'+ku=0
=4u''+4u'+101u=0
the initial conditions are given as u(0)=2 and u'(0)=8. Note that u' is velocity which is the derivative of position.
The problem states that the motion of the spring in underdamped, therefore, there will be complex roots
u(t)=C1e^(at)cos(bt)+C2e^(at)sin(bt), cmplex roots r=a+/-bi, r=(-1/2)+/-5i
u(t)=C1e^(-1/2t)cos(5t)+C2e^(-1/2t)sin(5t)
apply initial conditions to solve for C1 and C2:
u(0)=2=C1 C1=2
u'(0)=8=(2)(-1/2)+C2(5) C2=9/5
Therefore, x(t)= 2e^(-1/2t)cos(5t)+(9/5)e^(-1/2t)sin(5t)
Part 2:
c1= sqrt(2^2+(9/5)^2)=sqrt181/5
w1=sqrt(k/m)= sqrt101/2
p=-1/2 < from general solution u(t)
alpha= tan^(-1)(C2/C1)
Part three:
u(t)= 4u''+101u=0
r=0+/-sqrt101/2
u(t)=C1cos(sqrt101/2t)+C2sin(sqrt101/2)
u(0)=2=C1
C2=8/(sqrt101/2)
C0=sqrt(C1^2+C2^2)
w0=sqrt(101)/2
alpha0=tan^(-1)(C2/C1) <Be careful when determining the proper quadrant. Sometimes you need to add pi to this answer for these problems. I'm unsure as to why though.