Question

For the system below: a. Find the differential equation in terms of . b. Find the damped natural frequency (wa), and the damping ration (2). Leave the answer in terms of M, C, and L. C. For M = 0.026 lb-s-/in, L = 10 in, C = 0.0323 lb-s/in, 0(0) = 10 and 0(0) = 0, find the equation for the time response (t), and sketch a plot of this response. o to

Answer 1

By newton second law

free body diagram For Small deflection x = lo у = le r 7 hd i

(a)

ΣΤ = ΙΘ

-mgx – cil= 10

ալա = 1

m120 + cil+mgx = 0

ml + 16 + mgle = 0

(b) Damping factor

S = 2 km

P =3

m = min

k= mgl

cl2 2 ml2mgl 2m

Natural frequency

A = um

mgl Wn = V m2 = vi

Damped natural frequency

Wd = wnV1-52

(c) now putting values of m,c,l in equation in (a)

0.02Ꮾ · 102Ꮎ +0.0323 · 10Ꮎ +0.02Ꮾ 9.81 · 10Ꮎ = 0

2.6Ꮎ + 3.23Ꮎ + 2.55Ꮎ = 0

Solution of this differential equation will be

(t) = Cest

Where C is constant

now

e(t) = Csest

e(t) = Csest

put this value in equation

2.6. Csest +3.23. Csest + 2.55. Cest = 0

2.6.52 +3.23.8 +2.55 = 0

$1,2 = -0.62 +0.77i

e(t) = Ciel-0.62+0.77i)t + Coe(-0.62–0.771)

e(t) = -0.66+ (C (0.77i)t + Cel-0.77i)t)

e(t) = -0.66+ (Ci(cos(0.77t) +isin (0.77)) + C (cos(0.77t) – isin(0.77)))

e(t) = -0.66 (Alcos(0.77t) + A sin(0.77t))

where

A1 = C1+C2, A2 = iC-C2)

now apply boundary condition

(i)

Ꮎ(0) = 10

(0) = -0.66-(A cos(0.77-0) + A sin(0.77.0))

10 = A1 + A2.0

A1 = 10

e(t) = -0.66+ (10 · cos(0.77t) + A2 sin(0.77t))

(ii)  

0 = 40 = Ꮎ

er e-0.66+ (10.0.77-sin(0.77t) + A2.0.77cos(0.77t))

0 = -0.060 (10.0.77-sin(0.77.0) + A2.0.77cos(0.77 0))

A2 = 0

o(t) = -0.66t. 10. cos(0.77t)