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Use any software you like and provide code ! For small displacements, the force exerted by the elongation of a spring can be described by
Use any software you like and provide code !
For small displacements, the force exerted by the elongation of a spring can be described by Hooke's law: F = kx where F is the force, x is displacement from the equilibrium position, and k > 0 is a material dependent constant. For large displacements, and in general, springs can be divided between soft and hard, and a general nonlinear Hooke's law can be described as F = kx k123, where the _ case represents soft springs, and the + case represent hard ones. The following table contains noisy measurements of two different type of springs. Identify in both cases the parameters k and ki. Displacement 0.1 0.2 0.3 0.4 0.5 0.6 Force (Spring 1) 0.11723 0.16527 0.28996 0.355861 0.446146 0.476244 0.443929 0.424399 0.386496 Force (Spring 2) 0.00136968 0.0390108 0.0321574 0.0554252 0.0972051 0.128604 0.204889 0.296491 0.388957 0.7 0.8 0.9 For small displacements, the force exerted by the elongation of a spring can be described by Hooke's law: F = kx where F is the force, x is displacement from the equilibrium position, and k > 0 is a material dependent constant. For large displacements, and in general, springs can be divided between soft and hard, and a general nonlinear Hooke's law can be described as F = kx k123, where the _ case represents soft springs, and the + case represent hard ones. The following table contains noisy measurements of two different type of springs. Identify in both cases the parameters k and ki. Displacement 0.1 0.2 0.3 0.4 0.5 0.6 Force (Spring 1) 0.11723 0.16527 0.28996 0.355861 0.446146 0.476244 0.443929 0.424399 0.386496 Force (Spring 2) 0.00136968 0.0390108 0.0321574 0.0554252 0.0972051 0.128604 0.204889 0.296491 0.388957 0.7 0.8 0.9Step by Step Solution
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