A mass m=4 is attached to both a spring with spring constant k=325 and a dash-po

Question

A mass m=4 is attached to both a spring with spring constant k=325 and a dash-pot with damping constant c=4.

The mass is started in motion with initial position x0=1x0=1 and initial velocity v0=3v0=3 .

1.) 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)=C1eptcos(1t1)x(t)=C1eptcos(1t1). Determine C1C1, 11 ,11and pp.

2.) c1=

3.) w1=

4.) a1= (assume 0<a1<2pi)

5.) p=

Graph the function x(t)x(t) together with the "amplitude envelope" curves x=C1eptx=C1ept and x=C1eptx=C1ept.

Now assume the mass is set in motion with the same initial position and velocity, but with the dashpot disconnected ( so c=0c=0). Solve the resulting differential equation to find the position function u(t)u(t).
In this case the position function u(t)u(t) can be written as u(t)=C0cos(0t0)u(t)=C0cos(0t0). Determine C0C0, 00 and 00.

6.) c0=

7.) w0=

8.)a0= (assume 0<a0<2pi)

Explanation / Answer

This question is solved recently for different figures. Please have a look for understanding.

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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