9780134361307, Chapter 2, Problem 27P (1)

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Oct 30, 2023

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Chapter 2, Problem 27P Problem A bungee jumper weighing 160 Ib ties one end of an elastic rope of length 200 ft and stiffness. 10 Ib/in. to a bridge and the other end to himself and jumps from the bridge (Fig). Assuming the bridge to be rigid, determine the vibratory motion of the jumper about his static equilibrium position. Unstretched length, 200t FIGURE Step-by-step solution Step 1074 A bungee jumper is shown in the given figure: - e el Unstretched length, 200 ft Step 2074 Calculate the velocity (v) of the jumper as he falls from the 200 ft as follows: mgh= Lo <l v=2gh Here the mass of the jumper is m Substitute 386.4 in/s* for £ and 200 ft for } in the above equation to obtain the value of v as follows: v= ,2)((3864in/sz)x(ZOOXIZin) (1t=12in) =1361.881in/s Step 3074 The response of the jumper is given as: x(1)= 4, sin(w, +4,) Here the amplitude of the vibration is 4, the natural frequency of vibration is , and the phase angle of the vibration is g, Calculate the amplitude (4,) of the vibration as follows: fef) Here the displacement of the vibration is x, and the stiffness of the rope is k At the static equilibrium position = #(1=0) =1361.88in/s Substitute 101b/in for k., (%]Ib for m, 1361.88in/s for ¥, and the corresponding value of x, and X, in the above equation to obtain the value of A4, as follows: ' 1361.88 in/s ( 160 Jlb 386.4 ={o+ Ji01b/in (;604)"’ =(1361.881n/5)y[~3364)__ 101b/in =277.128in Thus the amplitude of the vibration is Step4 o7 Calculate the phase angle (¢,) of the vibration as follows: b= lan"[x"'fi] X =tan”'| 2 L 3 Vm Substitute the corresponding value of x, and , in the above equation to obtain the value of ¢, as follows: o 0 10 Ib/in f=tan”| ———— [ =— 1361.88 in/s =0° Thus the phase angle of the vibration is
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