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Question # 1: [25%] The system in the figure, m= 4 kg, a = 2 m, b= 3 m, k = 2500 N/m, c= 30 N.s/m, is subjected to a force ex

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Answer #1

Solution

Assuming small displacements we obtain Free body diagram as shown below.

3 F(t) caө таё kbe Inertia force

Taking anti clockwise rotation as poistive, we get,

mal – cao – khºa +bF(t) = 0

mai + cao + b^0 = bF(t)

Substituting the values we get,

4(2)28 +30(2) 6 + 2500(3) = (3)1000 sin wt

160 + 1200 + 225000 = 3000 sin wt

For steady state solution,let's assume

A= eiwt, F = Fewt , Fo= 3000

(-16w2 +i (120)w + 22500) Bewt = 3000et

3000 (22500 – 16w2) +i (120w)

10 = 3000 (22500 – 16w2)2 + (120w)? 500 – 16w2)

Therefore this is the expression for steady state amplitude when system is excited with \omega . Variation of steady state amplitude for various values of \omega can be observed from the plot shown below.

Variation of Steady state amplitude vs. Exciting frequency Steady state amplitude (in m) 10 15 20 25 Exciting frequency (in H

Natural frequency

(22500-16.2) + (120.1°=0

256w+ – 705600w2 + 225002 = 0

Take, \omega^2=\lambda

25632-7056000+ 22500 = 0

Solving this we get natural frequency ,

w = V1378 = 37.1214 rad/s

w = 37.1214/27 = 5.908 Hz

Displacements

a) Maximum displacement in the range 0 to 3 Hz occurs at 3 Hz

w = 3 x 27 = 18.85 rad/s

10 = = 3000 (22500 – 16(18.85)2)2 + (120(18.85))2

= 0.1768 m

b) Maximum displacement in the range 3 to 10 Hz occurs at natural frequency , i.e, 5.908 Hz

w = 3 x 27 = 37.1214 rad) s

3000 10 = = (22500 – 16(37.1214)2)2 + (120(37.1214))?

O = 0.67 m

c) Maximum displacement above 10 Hz occurs at 10 Hz

w = 10 x 27 = 62.8319 rad/s

10 = = 3000 (22500 – 16(62.8319)2)2 + (120(62.8319))2

O = 0.0725 m

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