Feynman Exercises Lecture Challenge Dale Lukas Peterson November 16, 2011
Restrictions on Solution Method In one of the Review lectures Feynman gave to his freshman students, just before their first big exam, he advised them as follows (copied from Feynman’s Tips on Physics, a problem-solving supplement to The Feynman Lectures on Physics): Now, all these things you can feel. You don’t have to feel them; you can work them out by making diagrams and calculations, but as problems get more and more difficult, and as you try to understand nature in more and more complicated situations, the more you can guess at, feel, and understand without actually calculating, the much better off you are! So that’s what you should practice doing on the various problems: when you have time somewhere, and you’re not worried about getting the answer for a quiz or something, look the problem over and see if you can understand the way it behaves, roughly, when you change some of the numbers. The challenge is to solve the problem given below (originally homework for FLP Vol. I, chapter 23) in the spirit of Feynman’s advice, above. It must be solved without using any calculus or differential equations or integral equations or difference equations, etc., without iterative numerical methods, nor any such fancy mathematical tricks! You may use only algebra, geometry, trigonometry, dimensional analysis, and Newtonian mechanics, in your solution, which should be guided by your physical intuition (however note: all intuitions used in solutions must be justified)! Your answer does not have to be exact, but it should at least be a very close approximation. And be sure to show all your work! Here is the problem:
Problem Statement The pivot point of a simple pendulum having a natural period of 1.00 second is moved laterally in a sinusoidal motion with an amplitude 1.00 cm and period 1.10 seconds. With what amplitude should the pendulum bob swing after a steady motion is attained?
Solution The first step in the solution process is to obtain the equation of motion. To this end, I introduce a few symbols to represent physical quantities: m, l, g, being the mass of the pendulum, the length from the pendulum pivot to the mass center, and the gravitational constant. Without loss of generality, I assume a simple point mass pendulum with a massless rod, and no dissipation of any kind. The velocity and acceleration of the pendulum mass relative to an inertial frame are ˆ v = x˙ˆi + θ˙ˆj × l sin θˆi + l cos θk ˆ = x˙ + lθ˙ cos θ ˆi + −lθ˙ sin θ k ˆ a = (¨ x + lθ¨ cos θ − lθ˙2 sin θ)ˆi + (−lθ¨ sin θ − lθ˙2 cos θ)k 1
x O
P ˆi
l
ˆ k
g
θ
m
ˆ × ˆi points out of the page. Figure 1: Pendulum with horizontally translating pivot. ˆj = k Two forces act on the mass; the gravitational force Fg and the force of the rod Fr : ˆ Fg = mg k
ˆ Fr = −F sin θˆi − F cos θk
where F is a function of time. Newton’s 2nd law yields: ˆ − F sin θˆi − F cos θk ˆ = m(¨ ˆ mg k x + lθ¨ cos θ − lθ˙2 sin θ)ˆi + m(−lθ¨ sin θ − lθ˙2 cos θ)k ˆ dividing through by ml and rearranging, yields: Dotting this equation with cos θˆi − sin θk, 0.01 2π 2 2π g ¨ sin t cos θ θ + sin θ = l l 1.1 1.1 Notice how dotting the Newton’s equation into the direction perpendicular to the rod is a convenient way to eliminate the constraint force Fr . Linearizing about the downwards position θ = 0, we have sin θ ≈ θ and cos θ ≈ 1, which yields the following second order linear differential equation: 0.01 2π 2 2π g sin t θ¨ + θ = l l 1.1 1.1 2π 2 2π Defining ωn2 = gl and f0 = 0.01 l 1.1 , and ω = 1.1 , we can rewrite the equation of motion as θ¨ + ωn2 θ = f0 sin ωt Assuming a particular solution of the form θp (t) = Θ sin ωt, differentiating twice with respect to time, and substituting into the equation of motion yields: (−ω 2 + ωn2 )Θ sin (ωt) = f0 sin ωt which implies that the amplitude of swing of the pendulum bob after steady motion is attained is: Θ= =
f0 2 ωn − ω 2 0.01 2π 2 l 1.1 2π 2 (2π)2 − ( 1.1 )
0.047619 l Where we made use of the fact that the natural frequency is related to the natural period of oscillation by ωn = 2π/T = 2π rad/s. Note that this solution depends on the length of the rod. However, if we assume Earth’s gravitational constant g = 9.81 m/s/s, then from g/l = ωn2 we can calculate that l = 0.2485 m. Under this assumption, the amplitude of oscillations is then: =
Θ = 0.1916 rad = 10.98 deg
This allows us to find the horizontal and vertical oscillations of X = l sin Θ + 11mm = 57.3mm l Y = (1 − cos Θ) 2 = 2.27mm
Alternate Method - Probably Cheating This problem cries out for a Lagrangian treatment. x = asin (ωt) + lsin (θ) y = −lcos (θ) ˙ x˙ = ωacos (ωt) + lθcos (θ) ˙ (θ) y˙ = lθsin Then L=T −V 2 2 m ˙ ˙ ωacos (ωt) + lθcos (θ) + lθsin (θ) = − mgy 2 m ˙ ω 2 a2 cos2 (ωt) + 2ωalθcos (ωt) cos (θ) + l2 θ˙2 + mglcos (θ) = 2 and ∂L ˙ = −mωalθcos (ωt) sin (θ) − mglsin (θ) ∂θ ∂L = mωalcos (ωt) cos (θ) + ml2 θ˙ ˙ ∂ θ d ∂L ˙ = −mω 2 alsin (ωt) cos (θ) − mωalθcos (ωt) sin (θ) + ml2 θ¨ dt ∂ θ˙ Now the Lagrangian equations of motion are ∂L ∂L = ˙ ∂θ ∂θ 2 2¨ ˙ ˙ −mω alsin (ωt) cos (θ) − mωalθcos (ωt) sin (θ) + ml θ = −mωalθcos (ωt) sin (θ) − mglsin (θ) 2 2¨ −mω alsin (ωt) cos (θ) + ml θ = −mglsin (θ) d dt
g ω2a θ¨ + sin (θ) = sin (ωt) cos (θ) l l ω2a θ¨ + ω02 sin (θ) = sin (ωt) cos (θ) l In the small angle approximation ω2a θ¨ + ω02 θ = sin (ωt) l With a steady state solution θ=
a ω2 sin (ωt) l ω02 − ω 2
Then (in the small angle approximation) x = asin (ωt) + lθ ω2 sin (ωt) = asin (ωt) + a 2 ω0 − ω 2 ω2 =a 1+ 2 sin (ωt) ω0 − ω 2 ω2 = a 2 0 2 sin (ωt) ω0 − ω ω02 where ω0 is the natural frequency of the pendulum, − ω2 ω is the driven frequency, and a is the amplitude of the driver.
and the amplitude of the driven oscillation is a
ω02
The Feynman Solution Start with the following assumptions 1. Assume the equilibrium solution for the x coordinate (small angle approximation) of the pendulum mass is of the form Asin (ωt), where ω is the driven frequency. 2. Assume that when equilibrium is achieved that the driver no longer supplies any energy to the system. Under these assumptions one can replace the pendulum with natural frequency ω0 with length l0 with one of length l that corresponds to natural frequency ω. The the problem of amplitude becomes one purely of geometry. First note that l = l0
ω02 . ω2
If ω < ω0 then l > l0 and the driver and pendulum motion are in phase, if ω > ω0 then l < l0 and the driven and pendulum motion are out of phase. Now consider figure 2 of an equivalent pendulum (below resonance). l is the length of the pendulum required for a natural resonance of ω. Then by geometry to get a driver amplitude of a we must have a A = l − l0 l l l − l0 ω2 l0 02 ω A=a ω02 l0 2 − l0 ω ω2 A=a 2 0 2 ω0 − ω A=a
which is the answer.
a
Plane of Driver l l0
A Figure 2: Equivalent Pendulum Below Resonance
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