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Ge 108 Homework #3 This homework is about the simple pendulum. In order to integrate the system numerically, we describe it as a pair of first-order differential equations, even though there is just one pendulum. This is not about the double pendulum. You can get a lot more information from the web, for example http://oldsite.vislab.usyd.edu.au/education/chaos/idjChaosLAB/node15.html http://www.cds.caltech.edu/~hinke/courses/CDS280/pendulum.html 1. Derive the equation of a pendulum that consists of a mass m hanging on a rigid massless rod of length L. The mass is constrained to move along the arc of a circle of radius L. The displacement x is the angle of the rod, in radians, relative to its hanging position at rest. The velocity y is the angular velocity dx/dt. There are two ways to derive the equations: (1) Distance along the arc of the circle is Lx, so acceleration along the arc is Ld2x/dt2 = Ldy/dt. The mass is m, and the component of gravity along the arc is mgsin(x). Use F= ma and you will get the equations below. (2) Angular displacement is x. Angular acceleration is d2x/dt2 = dy/dt; the moment of inertia is mL2; the torque due to gravity is mgLsin(x), and the equation is: moment of inertia times angular acceleration = torque. Either way, the equations are ! dx /dt = y, dy /dt = "(g /L)sin(x) + Fext Show that this reduces to the SHO equations when the angular displacement x is small and sin(x) = x. Then numerically integrate the equations with g/L = π2 so the period is 2.0. Let Fext = 0 (no external forcing) and start off with an initial displacement of 1.0 and velocity of zero. Plot x(t) and y(t). Find the analytic solution and compare with the numerical solution. 2. Repeat the exercise for the real pendulum, which has the sin(x) term as written above. Again set Fext = 0. What happens to the period when the initial x is not small, e.g., x(0) = 3π/4? Note that the initial value of |x| cannot be greater than π. Again plot x(t) and y(t). 3a. Again consider the real pendulum with the sin(x) term included. Start it with an initial velocity and no initial displacement. If the potential energy at the top is equal to the kinetic energy at the bottom, what is the minimum initial velocity at the bottom that will start the pendulum spinning around in a circle? Work this out analytically and then verify it numerically. Plot x(t) and y(t) when the pendulum goes around in a circle (i.e., over the top). 3b (This is an optional exercise, for those who feel comfortable with Mathematica; it’s OK to collaborate on this part). Plot y vs. x. Use modulo arithmetic (a built-in function), which subtracts off integer multiples of 2π so that the value of x is always between -π and +π. You should get a closed curve when the pendulum does not go over the top, and you should get a wavy curve when it does.4a. Linear pendulum with forcing and damping. Add an external force Fext = C cos(ωt). Resonance is when ω = π, since that is the natural frequency for the parameter values that we are using. You will need some damping to make the transients die away, so let the pendulum equation be ! dx /dt = y, dy /dt = "(g /L)sin(x) "#y + C cos($t) Since the natural frequency ω0 ≡ (g/L)1/2 = π, a good choice for the damping is γ = 0.5. This will alter the natural frequency by a small amount, and it will get rid of the transients after a few periods. For this part of the problem, replace sin(x) by x, so that you have the pure SHO. Calculate the amplitude of the oscillation (one-half the displacement from peak to trough) numerically for C = 1, after the transients have died away. Try it for several values of the forcing frequency on either side of the resonance, and compare the amplitude with the analytic solution. 4b (optional). Plot y vs. x as you did before. You should get an ellipse. In your y vs. x plot, try “strobing” at the forcing frequency, which means plotting the x-y position as a dot each time the forcing function repeats itself, e.g., at ωt = 0, 2π, 4π, 6π, …You should find, after the transients have died away, that all the points lie on top of each other − they repeat. This means the system is periodic at the forcing frequency. 5a. Repeat problem 4 for the real non-linear pendulum, which means keep the sin(x) term. There is no simple analytic solution, but you can examine the numerical solution for different values of the forcing amplitude C. First try C = 10 and ω = 0.5. Crank it up to C = 12 and you get chaos. Plot x(t) and y(t). The pattern doesn’t repeat. 5b (optional). When you plot y vs. x, you will not get a closed curve, signifying that the solution is not periodic. When you strobe it, the points will not lie on top of each other. It is impossible to predict exactly where the system will be many cycles into the future. I am not sure about this, but I believe the strobe plot is a “strange attractor,” a lacy filamentary structure with “holes” whose (x,y) locations are never visited. The second web site given above has a picture of what you get after a very long integration for the “parametrically forced damped pendulum.” That is a picture of a “strange


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CALTECH GE 108 - Homework 3

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