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Lecture

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Department
Physics
Course
PHY354H1
Professor
Erich Poppitz
Semester
Winter

Description
PHY254 Lecture 8 Oscillations in Phase Space and Dynamics ▯ Readings: For Oscillation dynamics: Morin 4.2 ▯ Simple Harmonic Motion Examples Simple Harmonic Motion ▯ Last time: We studied how to solve linear second order homogeneous ODEs. ▯ Standard example: Mass on a spring with no damping or driving forces: m + kx = 0 (1) ▯t ▯ tried a solution of the form xwhere A and ▯ are unknowns ▯ found: s k ▯ = ▯ ▯ m ▯ we learned nothing about A, but we now have 2 possible solutions for ▯. ODE theory says that a linear homogeous 2nd order ODE will have 2 linearly independent solutions. We’ve found both of these. The ▯nal solution is the superposition of these 2 solutions. So: h▯q k▯ i h ▯q k▯ i x(t) = Aexp ▯ m t + B exp ▯ ▯ m t . for some A and B are constants which can be determined by the initial conditions. ▯ For our speci▯c example, ▯k=m < 0. So we could right the exponential as: ▯▯q p ▯ ▯ ▯ q ▯ exp k=m ▯1 t = exp it k=m q Notice thatk=m appears in the exponential where the angular fre- quency should be. That makes it the angular frequency. Luckily for this example we know what the angular frequency should be (just from q preqious studies of springs) and it isk=m. So we can write ! = k=m and plug it into the solution. 1 ▯ There are a bunch of ways you can write the solution that are all equivalent: x(t) = Ae i!t+ Be ▯i!t x(t) = C cos!t + D sin!t x(t) = E cos(!t + ▯ ) 1 x(t) = F sin(!t + ▯ ) 2 i!t+3 x(t) = Re(Ge ) etc. The coe▯cients are determined by the initial conditions. Notice there are always 2 unknown constants in the above expressions. ▯ This type of motion that is purely sinusoidal is called \Simple Har- monic Motion" (SHM for short). ▯ Example: Using initial conditions to solve for constant: A particle on a spring is initially displaced a distance 0 from the origin and has an initial velocity v . If we write the solution as 0 x(t) = Ae i!t+ Be ▯i!t ▯nd the constants A and B. (You may have noticed that this form involves complex exponentials. In nothing we have done above did we say we only need the real component of the solution. But we know the solution must be real. The only way this can be is if the constants A and B themselves are complex and related in a speci▯c way. Lets solve for them and see what we get.) Initial conditions: At t = 0, x = x0and v = v .0 ▯ Answer: Lets write out the position and velocity for all time: i!t ▯i!t x(t) = Ae + Be _(t) = i!Ae i!t▯ i!Be ▯i!t Plugging in the initial conditions we ▯nd: x 0 A + B 2 v0= i!A ▯ i!B Solving for A in the ▯rst equation: A = x ▯ B and plugging it into 0 the second equation gives: v0= i!x 0 2i!B Solving for B gives: v0▯ i!x 0 B = ▯2i! Multiplying by i=i in order to make the denominator real gives us: (v ▯ i!x ) i 1 ▯ v ▯ B = 0 0 = x0+ 0i ▯2i! i 2 ! Plugging this into the equation for A gives: ▯ ▯ A = x ▯ B = 1 x ▯ v0 i 0 2 0 ! We have now solved for the constants of integration and found them to be complex. Although they are complex, they are related in a special i!t ▯i!t way. Because e and e are complex conjugates, so are the con- stants. This is the only way to make the solution for x real. Our ▯nal solution is: 1 ▯ v0 ▯ 1▯ v0▯ x(t) = x0▯ i e i!+ x0+ i e▯i!t (2) 2 ! 2 ! The answer we get above should be identical to what we would get if we had chosen any of the other forms for the solution above (e.g. x(t) = C cos!t + D sin!t). Lets verify that by solving the exact same problem with this representation and showing we get the same answer: ▯ Example: Using initial conditions to solve for constants: A particle on a spring is initially displaced a dista0ce x from the origin and has an initial velocity0v . If we write the solution as x(t) = C cos!t + D sin!t ▯nd the constants C and D. (Notice this is the exact same problem as the previous one). 3 ▯ Answer: Writing the position and velocity: x(t) = C cos!t + D sin!t x_(t) = ▯!C sin!t + !D cos!t Plugging in the initial conditions we ▯nd: x0= C v = !D 0 So C = x 0nd D = v =0. The solution is x(t) = x0cos!t + (v0=!)sin!t (3) Is equation (3) the same answer as equation (2)? Check by using DeMoivre’s theorem to write equation (2) in terms of sines and cosines: 1 ▯ v ▯ 1 ▯ v ▯ x(t) = x0▯ 0i ei!t+ x 0 0i e ▯i!t 2 ! 2 ! 1 ▯ v0 ▯ 1 ▯ v0 ▯ = x0▯ i (cos!t + isin!t) + x0+ i (cos!t ▯ isin!t) 2 ! 2 ! = x 0os!t + (v 0!)sin!t So yes they are the same. Notice that because the exponentials were complex conjugates and the constants in front of the exponentials were complex conjugates, all of the imaginary components cancel out of the ▯nal equation. ▯ In the previous lecture, we learned that the phase space curves of x = Acos(!t+▯) are ellipses. Thus, simple harmonic motion in phase space presents elliptical curves. 4 Studying the Simple Pendulum Example: Pendulum \down": Consider a pendulum
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