Problem Solving Session 10 Simple Harmonic Oscillator Solutions

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Problem Solving Session 10 Simple Harmonic Oscillator Solutions W13D3-0 Group Problem Gravitational Simple Harmonic Oscillator Two identical point-like objects each of mass m 1 are fixed in place and separated by a distance 2d. A third object of mass m 2 is free to move and lies on the perpendicular bisector of the line connecting the two objects at a distance y(t) above the midpoint of that line. The objects interact gravitationally. The force is attractive and has magnitude where r is the distance between the objects.! F grav = Gm 1m 2 r 2. a) Determine an expression for the y component of the gravitational force acting on the object in terms of m 1, G, y( t ), and d as needed using the fact that cos! = y / (d 2 + y 2 ) 1/ 2. b) Find an equation that the acceleration d 2 y / dt 2 of the object with mass m 2 satisfies in terms of the quantities m 1, G, y( t ), and d as needed. c) Use the approximation that y(t) << d in your equation in part a) to find an equation for d 2 y / dt 2 in terms of the quantities m 1, G, y( t ), and d as needed. d) Assume that at t = 0, the object of mass m 2 is at the equilibrium position but has an initial upward speed v 0. Write down an expression for the subsequent motion ( ) y t, in terms of m 1, G, v 0, and d as needed. Assume that y(t) << d.

2 Solution: The free body force diagram on the object of mass m 2 is shown in the figure below The y component of force acting on the object is F y =! Gm 1 m 2 (d 2 + y 2 ) 2cos". y From the diagram cos! = so the y component of force is (d 2 + y 2 1/ 2 ) F y =! 2Gm 1 m 2 y (d 2 + y 2 ) 3/ 2. Therefore Newton s Second Law yields the equation of motion m 2 d 2 y dt 2 =! 2Gm 1 m 2 y (d 2 + y 2 ) 3/ 2 When y(t) << d the equation of motion becomes d 2 y dt 2!! 2Gm 1 d 3 y. This is a simple harmonic oscillation equation of motion with solution where y(t) = y 0 cos( t) + v y,0 sin( t) = 2Gm 1 d 3. The initial conditions are y 0 = 0 and v y,0 = v 0 > 0. Therefore

3 y(t) = v 0 sin( t), v y (t) = v 0 cos( t).

4 IC_W13D3-3 Group Problem Kater s Pendulum Solutions A pendulum consists of a rod and two knife-edges separated from the center of mass l 1 and l 2 respectively (figure below left). The moment of inertia of the rod about the center of mass is I 0 = mk g 2 where k g is a constant called the radius of gyration. When the pendulum is pivoted about the upper knife-edge (figure below center), the period for small oscillations is T 1. When the pendulum is turned upside down and pivoted about the other edge (figure below right), and the period for small oscillations is T 2. The distances l 1 and l 2 are adjusted until T 1 = T 2! T. Express your answers to the following questions in terms of l 1, l 2, and T as needed. a) What is the radius of gyration k g? b) Using your results from part (a), show that the gravitational acceleration g can be determined by measuring the length between the knife-edges l 1 + l 2 and the period T.

5 Solution: We can treat each case as a physical pendulum. The torque equation of motion when suspended from the upper edge is!l 1 mg sin" = I 1 d 2 " dt 2. The moment on inertia about the upper knife-edge is I 1 = I 0 + ml 1 2 = m(k g 2 + l 1 2 ). Using the small angle approximation sin!!!. The torque equation becomes!l 1 mg" = m(k g 2 + l 1 2 ) d 2 " dt 2. Then the equation of motion for the pendulum is a simple harmonic oscillator equation of motion d 2! dt + l 1 g 2 (k 2 g + l 2 1 )! = 0. The period of oscillation is therefore T 1 = 2! (k g 2 + l 1 2 ) / l 1 g. When the pendulum is turned upside down and suspended from the other knife-edge and similar argument yields for the period T 2 = 2! (k g 2 + l 2 2 ) / l 2 g. Because distances l 1 and l 2 have been adjusted until T 1 = T 2! T, we can solve for k g by squaring the periods and setting them equal yielding 4!((k 2 g + l 2 1 ) / l 1 g) = 4!((k 2 g + l 2 2 ) / l 2 g) A little rearrangement yields k 2 g + l 2 1 = (k 2 g + l 2 2 )l 1 / l 2 or Therefore the period is k g 2 ((l 2! l 1 ) / l 2 ) = l 1 (l 2! l 1 ) k g 2 = l 1 l 2.

6 Hence T = 2! (l l + l 2 ) / l g = 2! (l 2 + l 1 ) / g g = 4! 2 (l 2 + l 1 ) / T 2. The answer only depends on the distance between the knife-edges l 1 + l 2, and the measured period, T.

7 W13D3-3 Group Problem Small Oscillations Lennard Jones Potential Solution A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones 6,12 potential U (r) = U 0 " # (r 0 / r) 12! 2(r 0 / r) 6 $ % ; r > 0. where r is the distance between the atoms. Let m denote the effective mass of the system of two atoms. Solution: a) Determine the value of r such that the potential energy is minimum. b) Determine the value of the second derivative of the potential energy at the minimum of the potential. c) Find the angular frequency of small oscillations about the stable equilibrium position for two identical atoms bound to each other by the Lennard-Jones interaction. The equilibrium points are found by setting the first derivative of the potential energy equal to zero, 0 = du dr = U "!12r 12 0 # 0 r! r 6 0 r!7 $ %. Therefore the equilibrium point is located at r = r 0. The second derivative of the potential energy is d 2 U dr = U " +(12)(13)r # 0 r!14! (12)(7)r 6 0 r!8 $ %. Evaluating this at r = r 0 yields d 2 U dr 2 (r 0 ) = 72U 0 r 0!2 The angular frequency of small oscillation is therefore = d 2 U dr (r ) / m = 72U / mr