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PHY354H1 (28)
Lecture

# L11_resonance.pdf

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Department
Physics
Course
PHY354H1
Professor
Erich Poppitz
Semester
Winter

Description
Lecture 11: Looking at Driven Oscillations and Resonance ▯ relevant reading: Morin 4.4 Oscillation Dynamics Recap I won’t go through this section in lecture, but I thought it might be nice to have a recap in your notes. The lecture will begin at the \Lecture Starts Here" section. So far, we have been studying F = mx  for a variety of forces. I recap the main equations and results below: 1. Restoring forces only ▯ Examples: Springs: F = ▯kx, Small angle pendulums: F = ▯mgx=l (where x is the angle)... ▯ Equation we solve: 2  + !0x = 0 where ! 0s the natural frequency of the oscillations and is given by the square root of the coe▯cient in front of the x in the forcing q divided by mass. For example, in the spring its ! =0 k=m, in q the pendulum its ! 0 g=l. ▯ Solution: x = Acos(! t0 + B sin(! t0 or equivalently written in other terms: x = C cos(! 0 + ▯ 1 i!0t ▯i!0t x = De + Ee i(0 t+2 ) x = Re[Ge ] where A;B;C;D;E;G;▯ and ▯1are ar2itrary constants that are determined by the initial conditions. 2. Restoring forces + Damping ▯ For our examples, we use the following damping force: F = ▯bx _ d 1 ▯ Equation we solve: 2  + 2 x_ + !0x = 0 where = b=2m ▯ The solution depends on the relative magnitudes of and ! . 0 There are 3 cases: (a) Underdamped: < ! 2 2: Then the solution is given by: 0 h i ▯ t ~t ▯i~t x = e Ae + Be q where !~ = ! 0 is real and so we get oscillations that decay exponentially. (b) Overdamped: > ! : Then the solution is given by: 0 h i x = e▯ t Ae t + Be ▯ t q 2 where = ▯ ! i0 real and < so we get exponential decay. (c) Critically damped: = ! : Then the solution is given by: 0 x = e ▯ t[A + Bt] so we get exponential decay (slightly faster than the over- damped case). 3. Restoring, Damping and Periodic Driving Forces ▯ We assume a simple sinusoidal driving force: F = F c0s(! t)d ▯ If we have a driving force consisting of a superposition of sinusoids, we can just add up our solutions for each sinusoid. ▯ Equation we solve: F 0  + 2 x_ + !0x = cos(!dt) m 2 ▯ Solution contains 2 parts added together: x = x +x where x is p h p the particular solution to the inhomogeneous equation and x ish the homogeneous solution to the homogeneous equation. For our cosine forcing above, the particular solution is: 2 2 F 0(! 0 ! )d F0 (2 ! d xp= 2 cos(!dt) + 2 sin(!dt) m R m R q where R = (! ▯ ! ) + (2 ! ) . 2 0 d d ▯ The homogeneous solution is one of the cases in item 2 above, depending on the relative size of and !0. ▯ Special Case: No damping: = 0. Although this is not physical (there is ALWAYS damping), its good to examine this as a limiting mathematical case. Plugging in = 0 into the equation for x p above we get: F 1 x = 0 cos(! t) p m (! ▯ ! ) 2 d 0 d and since = 0 < ! 2 the homogeneous solution is: 0 xh= C c1s(! t)0+ C si2(! t)0 where C 1nd C ar2 determined by the initial conditions. So that is the end of the recap. We are now going to start with this special case of no damping above to investigate driven systems a little closer. Lecture Starts Here Dependence of Solution on ! in No-damping Case ▯ Lets consider the \Special Case" of no damping in a driven system. The solution is given above. We will start by ignoring the homogeneous solution. The term in front of the cosine is: F0 1 C = 2 2 m (! 0 ! )d ▯ Notice that C is a function of ! dthe driving frequency). We can plot C as a function of !d: 3 2 2 2 2 ▯ Notice that C > 0 when ! < ! and C < 00when ! > ! . Also d 0 notice that there is an abrupt change in sign of C at ! = ! dnd th0t jCj becomes very large near ! . 0 ▯ This tendency for jCj to get very large when ! is nedr ! is know0 as a Resonance Phenomenon. ▯ Although there is nothing wrong with the above formulation, sometimes we like to represent solutions x spch that the constant in front of the cosine is ALWAYS positive. Notice that in the above case, C is not. But we can re-write x inpanother form such that the constant in front is always positive and represents the amplitude of the solution. That form is: x p Acos(! t ▯d▯) where A > 0 and ▯ is the phase. Remember that C changed sign at 2 2 !0. Our solution for ! 0and▯ = 0 m (! ▯0! ) d For ! > ! , we can use the following: d 0 ▯ ▯ F0 1 ▯F0 2 2 ▯ m (! ▯ ! ) 2 < 0 ) x = p ▯m (!0 ▯ ! d)▯cos(! d) 0 d Now, ▯cos(! t) = cos(! t + ▯) d d 4 (which can be proved using sum of angle formulas). This means we can satisfy the requirement that A > 0 by using: ▯ ▯ ▯F 0 1 ▯ A = jCj = ▯ 2 2 ▯ and▯ = ▯ ▯ m (! ▯0! ) d ▯ ▯ So a succinct way of writing the above requirements is: ▯ ▯ ▯ ▯ A = ▯F 0 1 ▯ for all! ▯m (! ▯ ! ) 2 ▯ d 0 d and 2 2 ▯ = 0 for! d! 0 ▯ We can plot A and ▯ as a function of ! (done using dython program driven.py): A and alph
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