.003 Fall 1999 Solution of Homework Aignment 4 1. Due to the application of a 1.0 Newton tep-force, the ytem ocillate at it damped natural frequency! d about the new equilibrium poition y k =. From the given Diplacement Repone plot, the equilibrium oet can be etimated to be = 4.4 x 10,4 meter. The damped natural frequency can be etimated by counting the cycle, 10, in.0 econd, which give a damped period T d =0: econd, and! d ==T d = 31:4 rad/ec. The logarithmic decrement LDR can be etimated by meauring the ratio of ucceive peak amplitude. Thi i generally a dicult meaurement tomake accurately. One way to increae the accuracy i to make a large number of meaurement baed on dierent pair of ucceive peak, and average the reult. Draw a line acro the plot at y =and meaure the amplitude of everal peak. Since only the ratio of ucceive peak will be ued, the meaurement can be in term of any convenient length unit. The meaurement below are in millimeter. Station Amplitude Ratio 0 6:, 1 4:86 0:781 4:00 0:83 3 3:08 0:770 4 :56 0:831 5 1:94 0:758 6 1:68 0:866 The average of thee rt ix ratio i 0.805, o the etimate for the LDR i ln 0:805 =,0:17. (a) The etimated tine i k = f a = 1:0 = 70 Newton/meter 4:4 10,4 (b) To etimate the ma m, when the tine k i known, it i neceary to know the undamped natural frequency! o. At thi point we know only the damped natural frequency! d. We can potpone etimation of the ma m until the damping ratio i etimated in (c) below, or we can aume, ince more than ten cycle of ocillation are viible, that the damping i uciently light to permit u to make the approximation that! o! d, in which cae m k! d = 70 =:30 kg (31:4) (c) To etimate the damping parameter b, we need to know the damping ratio. The damping ratio follow from the log decrement ratio, LDR, according to the formula = LDR + LDR
given in the Note accompanying Lecture 6. Inertion of the etimation LDR =,0:17 yield 0:069. (Note: With thi value of the undamped natural frequency i etimated a follow! p d 1, 31:4 q =31:5 rad/ec 1, (0:069) With thi etimate for! o the ma m, previouly etimated in (b) above at.30 kg, would now be etimated to be.9 kg. The damping parameter b i then etimated a b =! o m (0:069)(31:5)(:9) = 9:95 kg/ec, or 9.95 N/m/ec. With two plate-on-pring unit face-to-face, the eective tine of the combined ytem i k = 1000 N/m, and the eective damping parameter i b = 10 N/m/ec. For free vertical motion of the ma m =:0 kg, the diplacement y k from the equilibrium poition of the model atie the equation m d y k dt + bdy k dt + ky k =0 (a) The undamped natural frequency! o for the model i (b) The damping ratio for the model i k 1000 m = =:4 rad/ec, or f o =! o = (c) The decay time contant for the model i = 1! o = b! o m = 10 (:4)() =0:1116 1 (0:1116)(:4) =0:400ec =3:56 Hz 3. Initially the bungee jumper free-fall under the inuence of gravity and air reitance, then when the lack i taken up in the elatic cord, the cord exert an increaing retarding force which reduce her velocity to zero at level A, after which he bob up and down with decreaing amplitude until he come to ret at the equilibrium poition at level B. For the purpoe of a preliminary etimate of the condition up to intant at which level A i reached, neglect the eect of air reitance and damping in the cord. Under thi aumption energy i conerved. In the rt jump the equilibrium poition B i 0 feet below the point where the cord begin to tretch. The tine of the cord i then etimated a k = W = 150 0 =7:5pound/foot
The undamped natural frequency of the ma-pring ytem coniting of the jumper and cord i k k 7:5 m = W=g = =1:69 rad/ec 150=3: Let the elevation of the upper attachment point of the cord be denoted by h o, the elevation of the point where the lack i taken up and the cord begin to tretch be enoted by h 1, the elevation of the equilibrium point B be denoted by h B, and the elevation of the point A where the jumper' velocity rt vanihe be denoted by h A. (a) In the rt jump, h o, h 1 = 100 feet, and h o, h B = 10 feet. The level A can be located by equating the potential energy lot in the fall to the elatic energy in the cord. W (h o, h A )=k(h o, h A )= 1 k(h1,h A) = 1 k[(h o, h A ), 100] Thi reduce to a quadratic equation for (h o, h A ) whoe olution are (h o, h A ) = 10 66:3 feet The phyically ignicant rooti(h o,h A ) = 186:3 feet. The low point A i 66.3 feet below the equilibrium level B, or 186.3 feet below the upper attachment point. (b) The maximum downward acceleration i 3. feet/ec fall. during the initial free (c) The maximum upward acceleration occur at point A and i a max = (Max diplacement from equilibrium)! o = (66:3)(1:69) = 106:8ft/ec which i 3.3 time the acceleration of gravity. (d) The primary aumption made i that diipation of energy ha been neglected. The cord ha been aumed to behave like a linear pring. (e) In the econd jump the lack length i doubled which cut the tine in half, and double the ditance to 40 feet. The natural frequency i reduced to 0.897 rad/ec. The location of the new low point A' i obtained by olving a imilar quadratic equation with the new value of h 0, h 1 = 00 feet, h o, h B 0 = 40 feet, and = 40 feet. The reult i (h o, h A 0) = 40 13:7 feet The phyically ignicant rooti(h o,h A ) = 373 feet. The low point A i 13.7 feet below the equilibrium level B', or 373 feet below the upper attachment point.
(f) The level of the equilibrium point B' i 40 feet below the attachment point. (g) The maximum downward acceleration i till 3, feet/ec (h) The maximum upward acceleration i a max = (Max diplacement from equilibrium)! o = (13:7)(0:897) = 106:8ft/ec Although the jump involve a longer free fall and a greater extenion of the cord the maximum acceleration doe not change! 400 Bungee Jump 350 300 Level B Elevation [feet] 50 00 150 Level A Level B1 100 50 Level A1 0 0 1 3 4 5 6 Time [econd] Figure 1: Time Hitorie of Bungee Jump (i) The time hitorie of the two jump are hown in Fig.1. For plotting purpoe the elevation of the upper attachment point wa (arbitrarily) aigned the value 400 feet. The trajectory of free fall i indicated by the dahed-line parabola.
With 100 feet of lack the jumper' trajectory depart from the free-fall parabola at h = 300 ft and begin to ocillate about the nal equilibrium level B. With no damping the maximum excurion (and maximum acceleration) i at Level A. With 00 feet of lack the jumper' trajectory depart from the free-fall parabola at h = 00 ft and begin to ocillate about the nal equilibrium level B1. With no damping the maximum excurion (and maximum acceleration) i at Level A1. 4. Let x be the diplacement of the engine with repect to the tationary crate. The eective tine of the two end element i k, and the eective damping parameter i b. (a) The equation of motion for the engine i m d x dt +bdx dt +kx =0 (b) The engine weight i W = 500 pound. It ma i m = W=g. The value of k and b can be deduced from the given value of damped natural frequency! d =and damping ratio =0:707 from the equation! o =! d 1, = k W=g The undamped natural frequency i and b =! o W=g q 1, (0; 707) =8:88 rad/ec and k = W! o g The damping parameter i = (500)(8:88) (386) = 51: pound/inch W b =! o g =(0:707)(8:88)500 =8:13 pound/in/ec 386