Oscillation with damping

by songoku
Tags: damping, oscillation
songoku is offline
Dec10-12, 08:16 AM
P: 767
1. The problem statement, all variables and given/known data
Spring mass system with spring constant k = 40 N/m and mass 10 kg.
a. Find the angular speed and period. Draw the response X versus time t
b. Linear damping is added with ζ = 4 %. Find the angular speed and period. Draw the response
c. Viscous damping is added with c1 = 0.03. Find the angular speed and period. Draw the response
d. Another viscous damping with c2 = 0.015 added parallel to c1. Find the angular speed and period
e. If a force F = F0 cos (2.3t) is applied to the system, will it be closer to the resonance, undamped system or system with viscous damping?
f. The spring is divided into 4 parts of equal length and arranged as follows.

Find the angular speed and period of the system.

g. 2 viscous damper are added into the above system in series between point B and C with damping coefficient c1 = 0.03 and c2 = 0.02. Find the period and angular speed of this system

2. Relevant equations
kseries = k1 + k2

1/kparallel = 1/k1 + 1/k2

k' = (1 - ζ )k

[tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]

ω = √(k/m)

ω = 2π/T

3. The attempt at a solution
OK actually my teacher didn't teach anything in class. Only gave formula and homework, so I am just trying to use the formula without understanding the concept here because the test is tomorrow. Please excuse my poor understanding and for now I haven't drawn the response graph because I really don't understand how

a. ω = √(k/m) = √(40/10) = 2 rad/s

ω = 2π/T
T = π s

b. k' = (1 - ζ )k = (1 - 0.04) . 40 = 38.4 N/m
ω = √(k'/m) = 1.96 rad/s
T = 2π/ω = 3.2 s

c. [tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]
ω' = 2 √(1-(0.03/20)2) ≈ 1.99 rad/s
T = 2π/ω = 3.14 s

d. c = c1 + c2 = 0.04
[tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]

ω' ≈ 1.99 rad/s

T = 2π/ω = 3.14 s

e. no clue at all

f. because k is inversely proportional to length and the spring is divided into 4 parts, so the value of k for each part is 160 N/m

After some calculation, ktotal = 400 N/m

ω = √(k'/m) = 2√10 rad/s
T = 2π/ω = 0.99 s

g. do not understand at all

I am not sure whether my work right or wrong...

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