Solving the Harmonic Oscillator Equation NCSU Department of Math - PDF document

Solving the Harmonic Oscillator Equation NCSU Department of Math Morgan Root

Spring-Mass System � Consider a mass attached to a wall by means of a spring. Define y=0 to be the equilibrium position of the block. y(t) will be a measure of the displacement from this equilibrium at a given time. Take = = dy ( 0 ) y ( 0 ) y and v . 0 0 dt

Basic Physical Laws Newton’s Second Law of motion states tells us that the � acceleration of an object due to an applied force is in the direction of the force and inversely proportional to the mass being moved. This can be stated in the familiar form: � F net = ma In the one dimensional case this can be written as: � = & & F m y net

Relevant Forces = − � Hooke’s Law (k is F H ky called Hooke’s constant) = − � Friction is a force that & F F c y opposes motion. We assume a friction proportional to velocity.

Harmonic Oscillator Assuming there are no other forces acting on the system we have what is known as a Harmonic Oscillator or also known as the Spring-Mass- Dashpot. = + F F F net H F or = − − & & & m y ( t ) ky ( t ) c y ( t )

Solving the Simple Harmonic System + + = & & & m y ( t ) c y ( t ) ky ( t ) 0 If there is no friction, c=0, then we have an “Undamped System”, or a Simple Harmonic Oscillator. We will solve this first. + = m & & y ( t ) ky ( t ) 0

Simple Harmonic Oscillator = Notice that we can take K and look at the k m system : = − & & y ( t ) Ky ( t ) We know at least two functions that will solve this equation. = = y(t) sin ( K t) and y(t) cos ( K t)

Simple Harmonic Oscillator The general solution is a linear combinatio n of sin and cos. = ω + ω y ( t ) A cos( t ) B sin( t ) 0 0 ω = k Where . 0 m

Simple Harmonic Oscillator We can rewrite the solution as = Α ω + φ y ( t ) sin( t ) 0 ω = k where and 0 m ( ) Α = + φ = − ω 2 2 1 v y y ( ) and tan 0 0 0 ω v 0 0 0

Visualizing the System Undamped SHO 2 m=2 c=0 k=3 1.5 1 0.5 Displacement 0 -0.5 -1 -1.5 -2 0 2 4 6 8 10 12 14 16 18 20 Time

Alternate Method: The Characteristic Equation ω = Again we take and look at the system : k m 0 + ω = & & 2 y ( t ) y ( t ) 0 0 = rt Assume an exponentia l solution, y ( t ) e . ⇒ = = rt 2 rt & & & y ( t ) re and y ( t ) r e Subbing this into the equation w e have : + ω = 2 rt 2 rt r e e 0 0

Alternate Method: The Characteristic Equation To have a solution w e require that = − ω 2 2 r 0 ⇒ = ± ω r i 0 Thus we have two solutions ω − ω = = i i y ( t ) e and y ( t ) e 0 0

Alternate Method: The Characteristic Equation � As before, any linear combination of solutions is a solution, giving the general solution: ω − ω = + * i t * i t y ( t ) A e B e 0 0

Euler’s Identity � Recall that we have a ( ) ( ) ω = ω + ω i t e cos t i sin t relationship between e 0 0 0 and sine and cosine, and known as the Euler ( ) ( ) − ω = ω − ω i t e cos t i sin t 0 identity. 0 0 � Thus our two solutions are equivalent

Rewrite the solution We now have two solutions to : + = & & m y (t) ky(t) 0 One with complex exponentia ls ω − ω = + i t i t * * y ( t ) A e A e 0 0 and one with sine = Α ω + φ y ( t ) sin( t ) 0

Damped Systems � If friction is not zero then we cannot used the same solution. Again, we find the characteristic equation. = rt Assume y ( t ) e then, = = & rt & & 2 rt y ( t ) re and y ( t ) r e

Damped Systems Remember t hat we are now looking for a solution t o : + + = & & & y ( t ) C y ( t ) Ky ( t ) 0 & & & Subbing in y , y and y we have, + + = 2 rt rt rt r e Cre Ke 0 Which can only work if + + = 2 r Cr K 0

Damped Systems This gives us that : − ± − ω 2 2 = ω = C C 4 k r 0 where 0 2 m α = − ω 2 2 Let C 4 0 We have three options α > 1. 0 overdamped α = 2. 0 critically damped α < 3. 0 underdampe d

Underdamped Systems � The case that we are − ω < 2 2 C 4 0 interested in is the 0 underdamped system. The solution is − + α − − α C i C i = + y ( t ) Ae Be 2 2 or − = ω + ω C / 2 y ( t ) e ( A cos( t ) B sin( t )), ω = − = − 2 2 1 4 mk c 4 K C 2 2 m

Underdamped Systems We can rewrite this in a more intuitive form, − = Α c ω + φ t y ( t ) e sin( t ) 2 m − 2 4 mk c ω = Where , 2 m − Α = + φ = 2 2 1 A B , tan ( ), A B + c = = v y 0 0 A y and B 2 m ω 0

Visualizing Underdamped Systems Underdamped System 2 1.5 m=2 c=0.5 k=3 1 0.5 Displacement 0 -0.5 -1 -1.5 0 2 4 6 8 10 12 14 16 18 20 Time

Visualizing Underdamped Systems Underdamped Harmonic Oscillator 2 m=2 y(t)=Ae -(c/2m)t 1.5 c=0.5 k=3 1 0.5 Displacement 0 -0.5 -1 y(t)=-Ae -(c/2m)t -1.5 -2 0 2 4 6 8 10 12 14 16 18 20 Time

Visualizing Underdamped Systems Another Underdamped System 2 m=2 c=2 1.5 k=3 1 Displacement 0.5 0 -0.5 0 2 4 6 8 10 12 14 16 18 20 Time

Visualizing Underdamped Systems Underdamped Harmonic Oscillator 2.5 2 m=2 c=2 1.5 k=3 y(t)=Ae -(c/2m)t 1 0.5 Displacement 0 -0.5 -1 y(t)=-Ae -(c/2m)t -1.5 -2 -2.5 0 2 4 6 8 10 12 14 16 18 20 Time

Will this work for the beam? � The beam seems to fit the harmonic conditions. � Force is zero when displacement is zero � Restoring force increases with displacement � Vibration appears periodic � The key assumptions are � Restoring force is linear in displacement � Friction is linear in velocity

Writing as a First Order System Matlab does not work with second order equations � However, we can always rewrite a second order ODE as a � system of first order equations We can then have Matlab find a numerical solution to this � system

Writing as a First Order System Given this second order ODE + + = = = & & & & y ( t ) C y ( t ) Ky ( t ) 0 , with y ( 0 ) y and y ( 0 ) v 0 0 = = & We can let z ( t ) y ( t ) and z ( t ) y ( t ). 1 2 Clearly, = & z ( t ) z ( t ) 1 2 and = − − & z ( t ) Kz ( t ) Cz ( t ) 2 1 2

Writing as a First Order System Now we can rewrite the equation in matrix - vector form ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ & z 0 1 z = 1 1 ⎢ ⎥ ⎢ ⎥ ( t ) ⎢ ⎥ ( t ) − − & ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ z K C z 2 2 or ⎡ ⎤ 0 1 = = z(t) Az(t) ⎢ ⎥ where A − − ⎣ ⎦ K C with ⎡ ⎤ y = 0 ⎢ ⎥ z(0) ⎣ ⎦ v 0 This is now in a form that will work in Matlab.

Constants Are Not Independent � Notice that in all our solutions we never have c, m, or k alone. We always have c/m or k/m. � The solution for y(t) given (m,c,k) is the same as y(t) given ( α m, α c, α k). � Very important for the inverse problem

Summary � We can used Matlab to generate solutions to the harmonic oscillator � At first glance, it seems reasonable to model a vibrating beam � We don’t know the values of m, c, or k � Need the inverse problem