Question

The length of nylon rope from which a mountain climber is suspended has a force constant of 1.35 * 10⁴ N/m.

(a) What is the frequency at which he bounces, given his mass plus equipment to be 78.0 kg?

____________Hz

(b) How much would this rope stretch to break the climber's fall, if he free-falls 2.00 m before the rope runs out of slack?

____________m

(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

bounce frequency	_____________ Hz
distance stretched	______________m

Answer 1

(a) f = (1/2π)√(k/m) = (1/2π)√(1.35*10⁴kg/s² / 78kg) = 2.09 Hz

(b) PE gets converted into spring energy U. If the rope stretches "x," then
mg(2m + x) = ½kx²
78kg * 9.8m/s² * 2m + 78kg * 9.8m/s² * x = ½ * 13500N/m * x²
Dropping units for ease (x is in meters),
6750x² - 764.4 x - 1528.8 = 0
quadratic with roots at x = - value not posssible
and x = 0.535 m

(c)

Does that mean we're doubling up on the rope? Then we have two springs in parallel, and
keq = k1 + k2 = 2*k = 2.7e4 N/m
Then part (a) becomes f = 2.96 Hz

by repeating the above process

f = (1/2π)√(k/m) = (1/2π)√(2.7*10⁴kg/s² / 78kg) = 2.96 Hz
and (b) leads to
13500x² - 764.4 x - 1528.8 = 0
and x = 0.366 m