20131123

Physics quiz question: Hobble Creek Lodge human slingshot

Physics 205A Quiz 6, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Problem 10.57

"Devinsupertramp Roadtrip!!! Behind the Scenes"
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youtu.be/NzFo9N2DOYQ

A 70 kg person is attached to a bungee cord catapult[*] that can be approximated as a horizontal mass-spring system. The bungee cord is stretched 23 m from equilibrium, and when released, the person undergoes simple harmonic motion with a period of 4.0 s. Neglect friction and drag. The spring constant of the bungee cord is:
(A) 30 N/m.
(B) 60 N/m.
(C) 110 N/m.
(D) 170 N/m.

[*] Hobble Creek Lodge, 693 Hobble Creek Canyon Road, Springville, UT 84663.

Correct answer (highlight to unhide): (D)

The period T of a mass m attached to a spring with spring constant k is given by:

T = 2·π·√(m/k),

such that the spring constant k will be:

k = m·(2·π/T)2 = (70 kg)·(2·π/(4.0 s))2 = 172.718077 kg/s2,

or as expressed in more conventional units to two significant figures, the spring constant is 170 (kg·m/s2)·(1/m) = 170 N/m.

(Response (A) is m·g/A, where A = 23 m is the amplitude of the resulting idealized simple harmonic motion; response (B) is 2·m·g/A; and response (C) is m·(2·π/T).)

Sections 70854, 70855, 73320
Exam code: quiz06wR3k
(A) : 9 students
(B) : 10 students
(C) : 8 students
(D) : 37 students

Success level: 59%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70

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