damped spring system ?

Suppose you have a damped spring system

2x'' + 16x' +40x = 0 , x(0)=5, x'(0)=4

(a) Determine whether the system is over-, under-, or critically damped

(b) Find the motion x(t) when the damper is disconnected (2x'' +40x=0)

(c) Solve for the motion x(t) when the damper is connected and sketch the solution on the

same graph as the solution to (b), indicating how the damping changes the solution.

*I mainly need help with part c and the graph, not sure if I'm overdamping or underdamping in part a...

2 Answers

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  • 4 weeks ago
    Best answer

    2x’’ + 16x’ + 40x = 0, x(0) = 5, x’(0) = 4

    Given mx’’ + bx’ + kx = 0 where m > 0, b ≥ 0 and k > 0 

    if

    b² < 4mk ⎯⎯⎯⎯⎯➤( underdamping )

    b² > 4mk ⎯⎯⎯⎯⎯➤( overdamping )

    b² = 4mk ⎯⎯⎯⎯⎯➤( critical damping )

    then

    m = 2, b = 16 and k = 40 

    ∴ 

    b² = 16² = 256, 4mk = 4(2)(40) = 320

    ( a )

    since b² is less than 4mk, this is ‘underdamping’ .

    ( b )

    Assuming same initial conditions;

    2x’’+ 40x = 0

    2r² + 40 = 0

    r = ±2i√5

    x(t) = c₁ cos(2t√5) + c₂ sin(2t√5)

    5 = c₁ cos(0 * 2√5) + c₂ sin(0 * 2√5) ====> c₁ = 5

    x'(t) = -2√5 c₁ sin(2t√5) + 2√5 c₂ cos(2t√5)

    4 = -2√5 c₁ sin(0 * 2√5) + 2√5 c₂ cos(0 * 2√5) ===> c₂ = ⅖√5 

    x(t) = 5cos(2t√5) + ⅖√5 sin(2t√5)

    x(t) = Rcos( 2t√5 - ɸ ) where R = √( 5² + (⅖√5)² ) = ⅕√645 ≈ 5.08

    Rcos( 2t√5 - ɸ ) = 5cos(2t√5) + ⅖√5 sin(2t√5)

    Rcos( 2t√5)cos(ɸ) + Rsin( 2t√5)sin(ɸ) = 5cos(2t√5) + ⅖√5 sin(2t√5)

    Rcos(ɸ) = 5 and Rsin(ɸ) = ⅖√5 thus tan(ɸ) = (⅖√5)/5  

    ɸ = 0.177 rad (or 10.14° )

    x(t) ≈ 5.08cos( 2t√5 - 0.177 ) 

    ——————————————

    ( c ) 

    2x’’ + 16x’ + 40x = 0

    2r² + 16r + 40 = 0

    r = -4 ± 2i

    x(t) = e^(-4t) [ c₁ cos(2t) + c₂ sin(2t) ]

    5 = e^(-4*0) [ c₁ cos(2*0) + c₂ sin(2*0) ] ===> c₁ = 5

    x'(t) = e^(-4t)[-2c₁sin(2t) + 2c₂cos(2t)] - 4e^(-4t)[c₁cos(2t) + c₂sin(2t)]

    4 = e^(0)[ 2c₂ ] - 4e^(0)[ 5 ] thus c₂ = 12

    x(t) = e^(-4t) [ 5cos(2t) + 12sin(2t) ]

    x(t) = Re^(-4t) cos( 2t -  ɸ ) where R = √( 5² + 12² ) = 13

    Rcos(ɸ) = 5 and Rsin(ɸ) = 12 thus tan(ɸ) = 12/5 

    ɸ = 1.176 rad (or 67.38° )

    x(t) = 13e^(-4t) cos( 2t - 1.176 )

    ——————————————

    Attachment image
  • ted s
    Lv 7
    4 weeks ago

    [ 2 D² + 16 D + 40 ] = 2 [ ( D + 4)² + 4 ] ---> over damped

    x = c_1 e^(-4 t) cos 2t + c_2 e^(-4t) sin 2t....x(0) = 5 ---> c_1 = 5....

    x'(0) = 4 = 5 ( -4) + 4 c_2 ---> c_2 = 6

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