A mass m=4 is attached to both a spring with spring constant k=101 and a dash-po

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.

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