{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

newell_lec11-15_100713

# newell_lec11-15_100713 - MATH 254 Lectures 11-15 For the...

This preview shows pages 1–4. Sign up to view the full content.

MATH 254 Lectures 11-15 . For the next ten lectures, we will study 2 nd order linear ode’s d 2 y dt 2 + p ( t ) dy dt + q ( t ) y = g ( t ) (6.1) where, usually, p ( t ) ,q ( t ), and g ( t ) will be continuous functions of t on some open interval I . We will be particularly interested in the situation where p ( t ) and q ( t ) are constant and g ( t ) represents periodic forcing (e.g. g ( t )= A cos t ) and the applications of these equations to mass-spring systems and linear electrical circuits. For the next ±ve lectures, we will study 2 nd order linear homogeneous ode’s. We say (6.1) is homogeneous if g ( t ) 0, namely if d 2 y dt 2 + p ( t ) dy dt + q ( t ) y =0 . (6.2) The topics covered in lectures 11-15 are: How to write (6.1) as a system of two ±rst order ode’s The superposition property for homogeneous, linear ode’s Motivation and examples. The mass-spring The use of JOde for systems Method for solving 2 nd order, linear, homogeneous, constant coe cient ode’s Reduction of order A lesson on complex arithmetic Applications to free oscillations of mass-spring systems and electrical circuits ( § 4.9) The initial value problems for (6.1), (6.2) Principle of superposition and the Wronskian Linear dependence (ld) and linear independence (li) of functions on an interval The connection between ld, li and Wronskians References: NSS 4.1-5, 4.9 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
How to write (6.1) as a system of two Frst order ode’s. Let dy dt = v. Then dv dt = d 2 y dt 2 = p ( t ) v q ( t ) y + g ( t ) So the ode (6.1) is equivalent to the system dy dt = v (6.3a) dv dt = p ( t ) v q ( t ) y + g ( t ) (6.3b) of two Frst order ode’s. Remark: There are many ways we could have split (6.1). ±or example, we could have deFned v as dy dt +2 y .Th en dv dt = d 2 y dt 2 dy dt =( p ( t ) + 2) v q ( t ) y + g ( t ). The choice of how to split is usually dictated by physical considerations. ±or exam- ple, we will next derive the equation for a mass-spring system and in that case it is natural to choose as the two dependent variables the displacement x ( t ) and the velocity u ( t )= dx ( t ) dt of the mass. Motivation Example: Mass-Spring System. > > > > > > > > > > > > > > > mg > > > > > > > > > > > > > > > mg ABC l 0 l x ( t ) ------------ ------------- -------- tension kl tension k ( x + l ) 2
Facts: A spring (very light) hangs under its own weight. When stretched or com- pressed, it develops a tension proportional to the amount of stretching or compres- sion which seeks to restore the spring to its equilibrium length l 0 . In B: The weight mg stretches the spring by a length l . The force that balances mg is the tension kl .Thecoe cient k is known, so l is found by mg = . Note: The law that says the tension in the spring is proportional to the extension is sometimes known as Hooke’s law and k is sometimes called the spring constant or Hooke’s constant. It is also related to the Young’s modulus of the elastic spring.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 27

newell_lec11-15_100713 - MATH 254 Lectures 11-15 For the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online