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1. Jumping on a trampoline can be modelled by a mass-spring system. A person is jumping on a trampoline vertically and lightly so that his/her feet do not leave the trampoline. The jumper has a mass of m = 50 kg. The natural height of the trampoline is 1.5 m above ground level and the trampoline has a spring constant k = 1000 Nm¯¹. At equilibrium the trampoline is compressed by a distance s metres. Natural height System at equilibrium 1.5m XI T H SM 1000 1.5m Air resistance acts on the jumper with a damping constant 3 = 50 Nsm-¹. Assume that the gravitational constant is g = 10 ms ². Let y(t) be the distance in metres of the jumper below the equilibrium position at time t seconds after the start of the jump, and let h(t) be the height in metres of the jumper above ground level at time t. At time t = 0, the jumper starts at height 0.6 m above the ground and the velocity of the jumper is 0. You may assume that the equation of motion for the system is

1. Jumping on a trampoline can be modelled by a mass-spring system. A person is jumping on a trampoline vertically and lightly so that his/her feet do not leave the trampoline. The jumper has a mass of m = 50 kg. The natural height of the trampoline is 1.5 m above ground level and the trampoline has a spring constant k = 1000 Nm¯¹. At equilibrium the trampoline is compressed by a distance s metres. Natural height System at equilibrium 1.5m XI T H SM 1000 1.5m Air resistance acts on the jumper with a damping constant 3 = 50 Nsm-¹. Assume that the gravitational constant is g = 10 ms ². Let y(t) be the distance in metres of the jumper below the equilibrium position at time t seconds after the start of the jump, and let h(t) be the height in metres of the jumper above ground level at time t. At time t = 0, the jumper starts at height 0.6 m above the ground and the velocity of the jumper is 0. You may assume that the equation of motion for the system is

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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(a) Draw a diagram of the system when the jumper is below the equilibrium position and moving up. Show all forces acting on the jumper and label them. (b) Find s, and hence relate h(t) to y(t). (c) Express the initial conditions in terms of y and y. (d) Solve for y(t). (e) Sketch h(t) versus t.
Transcribed Image Text:(a) Draw a diagram of the system when the jumper is below the equilibrium position and moving up. Show all forces acting on the jumper and label them. (b) Find s, and hence relate h(t) to y(t). (c) Express the initial conditions in terms of y and y. (d) Solve for y(t). (e) Sketch h(t) versus t.
1. Jumping on a trampoline can be modelled by a mass-spring system. A person is jumping on a trampoline vertically and lightly so that his/her feet do not leave the trampoline. off The jumper has a mass of m = 50 kg. The natural height of the trampoline is 1.5 m above ground level and the trampoline has a spring constant k = 1000 Nm¯¹. At equilibrium the trampoline is compressed by a distance s metres. Natural height System at equilibrium off 1.5m It T Air resistance acts on the jumper with a damping constant 3 = 50 N sm-¹. Assume that the gravitational constant is g = 10 ms ². Sm 1.5m Let y(t) be the distance in metres of the jumper below the equilibrium position at time t seconds after the start of the jump, and let h(t) be the height in metres of the jumper above ground level at time t. At time t = 0, the jumper starts at height 0.6 m above the ground and the velocity of the jumper is 0. You may assume that the equation of motion for the system is 50y + 50y + 1000y = 0. Please turn over
Transcribed Image Text:1. Jumping on a trampoline can be modelled by a mass-spring system. A person is jumping on a trampoline vertically and lightly so that his/her feet do not leave the trampoline. off The jumper has a mass of m = 50 kg. The natural height of the trampoline is 1.5 m above ground level and the trampoline has a spring constant k = 1000 Nm¯¹. At equilibrium the trampoline is compressed by a distance s metres. Natural height System at equilibrium off 1.5m It T Air resistance acts on the jumper with a damping constant 3 = 50 N sm-¹. Assume that the gravitational constant is g = 10 ms ². Sm 1.5m Let y(t) be the distance in metres of the jumper below the equilibrium position at time t seconds after the start of the jump, and let h(t) be the height in metres of the jumper above ground level at time t. At time t = 0, the jumper starts at height 0.6 m above the ground and the velocity of the jumper is 0. You may assume that the equation of motion for the system is 50y + 50y + 1000y = 0. Please turn over
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