Newton's second law F=ma is a generator of ODES. If you have a function that models your force, you replace the F with that function and there you have your ODE. My go-to example for explaining this is Hooke's Law. Suppose you have a spring that is on the x- axis where the left end of the spring is fixed so it cannot move. Suppose further that x=0 coincides with the right end of the spring (the end that can move). Robert Hooke deduced that if you move the right end of the spring from x=0 to some other x then the spring will resist you with a force given by: F = -kx. That's Hooke's Law. If you bring Hooke's Law together with Newton's second law you get this: ma = -kx remembering that a = m d'r = -kx dt² we get that There's a second order ODE. Now here's your problem: You encounter the first ever non-Hookeian spring. That is, you find a spring that does NOT obey Hooke's Law. Instead, the spring obeys a new law developed by Professor Robert Schmitty. Here is Schmitty's Law: "A Schmitty Spring exerts force that is proportional to its velocity and in the opposite direction." Suppose your Schmitty spring has a Schmitty spring constant of k=3 and is attached to a mass of 6kg. Further suppose that you stretch your Schmitty spring in the positive direction at a rate of 2 meters per second and release it when it is stretched 1 meter past its equilibrium length. Find an ODE that models the behavior of this Schmitty spring system, classify this ODE, then solve the ODE to find a function x(t) the gives you the position of the mass at time t. Some helpful hints: • Suppose the mass at the end of the spring is starting at position x=0 • Even though this is second order, you've already done problems of this type in chapter 1. I won't say where exactly but I'm sure you can find them. • Professor Robert Schmitty is a fictional character and I hope Schmitty springs are fictional as well.
Newton's second law F=ma is a generator of ODES. If you have a function that models your force, you replace the F with that function and there you have your ODE. My go-to example for explaining this is Hooke's Law. Suppose you have a spring that is on the x- axis where the left end of the spring is fixed so it cannot move. Suppose further that x=0 coincides with the right end of the spring (the end that can move). Robert Hooke deduced that if you move the right end of the spring from x=0 to some other x then the spring will resist you with a force given by: F = -kx. That's Hooke's Law. If you bring Hooke's Law together with Newton's second law you get this: ma = -kx remembering that a = m d'r = -kx dt² we get that There's a second order ODE. Now here's your problem: You encounter the first ever non-Hookeian spring. That is, you find a spring that does NOT obey Hooke's Law. Instead, the spring obeys a new law developed by Professor Robert Schmitty. Here is Schmitty's Law: "A Schmitty Spring exerts force that is proportional to its velocity and in the opposite direction." Suppose your Schmitty spring has a Schmitty spring constant of k=3 and is attached to a mass of 6kg. Further suppose that you stretch your Schmitty spring in the positive direction at a rate of 2 meters per second and release it when it is stretched 1 meter past its equilibrium length. Find an ODE that models the behavior of this Schmitty spring system, classify this ODE, then solve the ODE to find a function x(t) the gives you the position of the mass at time t. Some helpful hints: • Suppose the mass at the end of the spring is starting at position x=0 • Even though this is second order, you've already done problems of this type in chapter 1. I won't say where exactly but I'm sure you can find them. • Professor Robert Schmitty is a fictional character and I hope Schmitty springs are fictional as well.
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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