Question

Class Management Help HW10: Oscillatory motion and waves (Chapter 16) Begin Date: 10/31/2019 3:00:00 PM - Due Date: 11/20/2019 11:59:00 PM End Date: 12/18/2019 11:59:00 PM (0%) Problem 7: The length of nylon rope from which a mountain climber is suspended has a force constant of 1.55 x 10' N/m. Randomized Variables km 1.55 x 10 N/m theexpert.com-triking id: 51-ES-D-A685-17564. In scordance with Expert TA Terms of Service copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account 255 Part(a) What is the frequency, in hertz, at which he bounces given his mass and the mass of his equipment is 98 kg? f=1.995536 1.996 ✓ Correct! 25% Part (b) How far, in centimeters, would this rope stretch to break the climber's full if he free-falls 2.6 m before the rope runs out of slack? 13.539109 13.54 X Incorrect! Give Up used. x 255 Part(e) Assume that tying two ropes of the same length together would have the same effect on the spring constant as two springs connected in series. What would the frequency of oscillation be, in hertz, if the rope was twice as long? 255 Part (d) If the climber falls the same distance before running out of slack, how far would the rope stretch, in centimeters, if it was twice as long? Grade Summary Deductions
PART C AND PART D

Answer 1

K= 1.55xrotrilm mc 98 kg. Part 6 h = 2.6m ,n=? hoht x = 12.6+) m - by the conservation of energes. PE = Upring. mg h = of the mg (2.64*) = of lem? 9879.81 (7.57%) = 5x11.55x1044 jy? 2500 + 966,44 = 7750 a 77503 -961.4% - 2500 =0. -IV 6=14403 29 -6-960.43 1.16961.4)*-(487750 X(-25cv)] X - &x7750 = 0.6336 m ;x=-5093 m X (-re) X x = 63.36x10²m = 63.36 m

Part (c) » twoo Springs are connected, Here fore the total spring constant is ke=k, tkr 11, 2k, = 11:55 x15t) Nim 16e = 16;+ky = 11:55X0 +(1.55 416) Hon Ke= 3.1x104NIM New frequenas trw=288, Se ne pa- zure her - od Belgien H= 2.83.42 Postelsame as part by PE = Vepsing. ho = (bo +.) – (26 +*2) mgh = $k x² mg (4.64X3) = $13.14194) 98X9.81 (7.6+43) = $(3.18043> 7500 +961.42€ 15500 ML 155003; -961-4X - 250 = 0 -6901.41 = (-qbi9y(4 X15500 XC-15CO)] _-/224ae 287-0.372 Typ za 2X15500 =0.4338 m @ r es xifres X 0.437&m= 43. 38X m * = 43.38 cm