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A Damping Problem 

? 2008 Waterloo Maple Inc 

 

Introduction 

 

The system consists of a machine containing a large rotating mass. This is to be placed on an elastic material. The damping, , is to be determined by visualising the damped amplitude as a function of frequency and damping. 

 

Image 

 

 

Parameters of the System 

 

The machine has a mass `#msub(mi( = 1000 Unit('kg'), and the rotating mass is `#msub(mi( = 500Unit('kg'). The elastic material has spring rate d = 100000Unit(`/`(`*`('N'), `*`('m')))   . Let r be the radius of the crankshaft and ωt  its angular velocity. 

`:=`(val, `#msub(mi( 

 

Solution 

 

The oscillation of the machine is described by the following differential equation, where c describes the damping coefficient: 

 

`:=`(deq, `+`(`*`(`+`(`#msub(mi( 

`+`(`*`(`+`(`#msub(mi( (3.1)
 

Solving this gives: 

 

`:=`(sol, dsolve(deq, x(t))); 1 

x(t) = `+`(`*`(exp(`+`(`/`(`*`(`/`(1, 2), `*`(`+`(`-`(c), `*`(`^`(`+`(`*`(`^`(c, 2)), `-`(`*`(4, `*`(d, `*`(`#msub(mi(
x(t) = `+`(`*`(exp(`+`(`/`(`*`(`/`(1, 2), `*`(`+`(`-`(c), `*`(`^`(`+`(`*`(`^`(c, 2)), `-`(`*`(4, `*`(d, `*`(`#msub(mi(
(3.2)
 

Since we are only interested in the stationary state, the exponential functions (which describe the transient effect) can be ignored. 

 

`:=`(x_stat, unapply(subs(_C1 = 0, _C2 = 0, op(2, sol)), t)); 1 

proc (t) options operator, arrow; `+`(`-`(`/`(`*`(`+`(`*`(`+`(`*`(`+`(`-`(`#msub(mi( (3.3)
 

The force exerted on the base is: 

`:=`(`#msub(mi( 

`+`(`-`(`/`(`*`(d, `*`(`+`(`*`(`+`(`*`(`+`(`-`(`#msub(mi(
`+`(`-`(`/`(`*`(d, `*`(`+`(`*`(`+`(`*`(`+`(`-`(`#msub(mi(
(3.4)
 

 

To determine the amplitude, we set the oscillating part to its maximal value, using the following addition theorem: 

 

 

`+`(`*`(a, `*`(sin(`*`(omega, `*`(t))))), `*`(b, `*`(cos(`*`(omega, `*`(t)))))) = `*`(`^`(`+`(`*`(`^`(a, 2)), `*`(`^`(b, 2))), `/`(1, 2)), `*`(sin(`+`(`*`(omega, `*`(t)), arcsin(`/`(`*`(b), `*`(`^`(`+... (3.5)
 

match(subs(`*`(omega, `*`(t)) = ttt, `#msub(mi( 

true (3.6)
 

 

{a = `+`(`-`(`/`(`*`(c, `*`(`^`(omega, 5), `*`(r, `*`(`#msub(mi(
{a = `+`(`-`(`/`(`*`(c, `*`(`^`(omega, 5), `*`(r, `*`(`#msub(mi(
(3.7)
 

 

From this, the damped amplitude is given by : 

 

`:=`(`#msub(mi( 

proc (omega) options operator, arrow; `/`(`*`(`^`(omega, 2), `*`(r, `*`(`#msub(mi( (3.8)
 

 

The unbalancing force (without damping) is: 

 

`:=`(`#msub(mi( 

proc (t) options operator, arrow; `*`(`^`(omega, 2), `*`(r, `*`(`#msub(mi( (3.9)
 

Therefore the unbalancing amplitude is: 

 

`:=`(`#msub(mi( 

proc (omega) options operator, arrow; `*`(`^`(omega, 2), `*`(r, `*`(`#msub(mi( (3.10)
 

The quality of the damping is determined by the ratio of the amplitudes,  

 

`:=`(A, unapply(normal(`/`(`*`(`#msub(mi( 

proc (omega, c) options operator, arrow; `/`(`*`(`^`(`+`(`*`(`^`(d, 2)), `*`(`^`(c, 2), `*`(`^`(omega, 2)))), `/`(1, 2))), `*`(`^`(`+`(`*`(`^`(omega, 4), `*`(`^`(`#msub(mi( (3.11)
 

 

We determine the maximum amplitude with respect to c: 

 

`:=`(tmp1, [solve(diff(A(omega, c), omega) = 0, omega)]) 

[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
[0, `/`(`*`(`^`(`*`(`+`(`#msub(mi(
(3.12)
 

 

Choosing the positive root, which corresponds to the physical reality: 

 

`:=`(tmp2, select(proc (x) options operator, arrow; evalf(`<`(0, Re(subs(val, c = 1.0, x)))) end proc, tmp1)[1]); 1 

`/`(`*`(`^`(`*`(`+`(`#msub(mi( (3.13)
 

 

 

 

`:=`(tmp, unapply(subs(val, tmp2), c)); 1 

proc (c) options operator, arrow; `+`(`/`(`*`(`/`(1, 1500), `*`(`^`(1500, `/`(1, 2)), `*`(`^`(`+`(`-`(15000000000000), `*`(100000, `*`(`^`(`+`(22500000000000000, `*`(300000000, `*`(`^`(c, 2)))), `/`(1... (3.14)
 

 

`:=`(Max_A, unapply(simplify(subs(val, A(tmp(c), c))), c)); 1 

proc (c) options operator, arrow; `/`(`*`(`^`(`+`(225000000, `*`(3, `*`(`^`(c, 2)))), `/`(1, 4))), `*`(`^`(`/`(`*`(`+`(675000000000000000000, `-`(`*`(45000000000000000, `*`(`^`(`+`(225000000, `*`(3, `...
proc (c) options operator, arrow; `/`(`*`(`^`(`+`(225000000, `*`(3, `*`(`^`(c, 2)))), `/`(1, 4))), `*`(`^`(`/`(`*`(`+`(675000000000000000000, `-`(`*`(45000000000000000, `*`(`^`(`+`(225000000, `*`(3, `...
proc (c) options operator, arrow; `/`(`*`(`^`(`+`(225000000, `*`(3, `*`(`^`(c, 2)))), `/`(1, 4))), `*`(`^`(`/`(`*`(`+`(675000000000000000000, `-`(`*`(45000000000000000, `*`(`^`(`+`(225000000, `*`(3, `...
(3.15)
 

 

Graphical Display of the Solution 

 

We insert the parameters into the formula for the amplitude and display it. 

 

 

Button#Parameter Insertion & Plot 

 

Conclusion 

 

Maple enables us to solve the differental equation exactly. With this symbolic solution we could calculate and visualise the ratio of the amplitude of the oscillation to the amplitude delivered to the base. This ratio is important for the quality of the damping. 

 

Reference 

Heinz Waller and Reinhard Schmidt, Schwingungslehre f?r Ingenieure, Wissenschaftsverlag.