During testing the suspension system of a 2000kg automobile compresses 12cm unde

Question

During testing the suspension system of a 2000kg automobile compresses 12cm under a 20,000N force. The resulting oscillation amplitude decreases by 50% each cycle. What are the values of the spring constad k, and the damping constant B, for one of the four shock absorber systems on the car. Assume distribution of forces across the four spring- shock sysem of car.

* What is the Total Mechanical energy stored in the oscillation when it begins?

*How much energy will the system have on its first pass through the equilibbrium position?

*What will be the speed of the system at the first pass through equilibrium?

Thanks for any help, I will rate.

Explanation / Answer

Part a)

For this problem we will use the expression for a damped harmonic oscillator

w2= wo2 –(b/2m)2

w02= k/m

k calculated with the law of Kooke

F= -k x

k = F / x

k = 20000/0.12

k = 1.666 105 N/m

Wo2 = 1.666 105 /2000

Wo2 = 8.333 101 rad/s

Wo2 = 83.333 rad/s

in the first cycle is W

in the second cycle is W / 2

w2= wo2 –(b/2m)2   (1)

(w/2)2= wo2 –(b/2m)2

w2= 4 (wo2 –(b/2m)2) (2)

equate 1 y 2

wo2 –(b/2m)2= 4 (wo2 –(b/2m)2)

- (b/2m)2+ 4(b/2m)2= 4wo2 –w02

3 b2/4m2= 3 w02

b2= wo24m

b2 = 83.333 4 2000

b = 816.49

this is the total damping factor, if this is distributed between four buffers, b is reduced to ¼

Part b)

Initial mechanical energy is

Em = ½ k A2

Em = ½ 1.6666 105 0.122

Em = 1.2 103 J

Part c)

Em = K = ½ m v2

Em = ½ 2000 1.0572

Em = 1.117.25 J

Part d)

V = dx/dt

p = b/2m

x = A Cos (Wt) e -p t

V = A (- W Sin (Wt)e-pt -p e-p t Cos(wt) )

n the equilibrium position X = 0

when the function Cos = 0 soWt = /2

t = /2W

V = A (- W Sin (W /2W)e-pt -p e-p t Cos(w/2W) )

V = A (- W Sin ( /2)e-pt -p e-p t Cos(/2) )

sin /2 =1

Cos /2 =0

V = -AW e-p/2w

in the first oscillation W = W0

V = - 0.12 sqrt(83.333) e-u

u= p /(2W0)

u = 816.44/ (2 2000) / ( 2 sqrt 83.333)

u= 0.035

e-0.035 = 0.965

V = -1.057 m/s

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