Ejercicio Exercise 1 · Física II URV 2024

Tema: Oscillations: simple harmonic, damped and forced · Del examen Exámenes de Física II: Teoría y Laboratorio (Mayo 2024) (2024)

Enunciado

Exercise 1 (10 points). Bungee jumper oscillation

An individual of mass m=60 kgm = 60\text{ kg} leaps from a bridge situated at a height h=100 mh = 100\text{ m} above the ground, secured to an elastic rope of natural length L0=20 mL_0 = 20\text{ m} and spring constant kk. The jumper experiences free fall until the rope reaches its unstretched length. Beyond this point, the cord stretches and applies an upward force that decelerates the jumper until he momentarily halts and rebounds into oscillatory motion. Assume air resistance and the mass of the rope are negligible (g=9.8 m/s2g = 9.8\text{ m/s}^2).

a.

Calculate the spring constant kk required so that the jumper stops (v=0v = 0) just before reaching the ground. [2 points]

b.

Determine the maximum elongation of this same rope if a second jumper of mass m2=90 kgm_2 = 90\text{ kg} performs the jump. [2 points]

c.

Compute the speed of this second jumper when reaching the ground level. [2 points]

d.

Establish the differential equation and solve for the equation of motion x(t)x(t) of this second jumper once the rope begins to stretch, setting x(0)=0x(0) = 0 at the moment the rope starts extending. [4 points]

Enunciado: Ejercicio Exercise 1 · Física II URV 2024

Solución

Según los criterios oficiales de corrección

a. Calculate the spring constant k required so that the jumper stops (v = 0) just before reaching the ground.

Solución: 18.37518.375 N/m

b. Determine the maximum elongation of this same rope if a second jumper of mass m2 = 90 kg performs the jump.

Solución: 113.0113.0 m

c. Compute the speed of this second jumper when reaching the ground level.

Solución: 25.5625.56 m/s

d. Establish the differential equation and solve for the equation of motion x(t) of this second jumper once the rope begins to stretch, setting x(0) = 0 at the moment the rope starts extending.

Solución: 48−48cos⁡(0.452t)+43.82sin⁡(0.452t)48 - 48\cos(0.452t) + 43.82\sin(0.452t)

La explicación paso a paso, hecha con IA, es Premium. Premium

Resuélvelo aquí mismo: corrección automática, pistas y solución paso a paso.

‹ Ejercicio Lab 3 · 2024 · Ejercicio Exercise 1 - Simple harmonic motion · 2024 ›

Más ejercicios de Oscillations: simple harmonic, damped and forced