ps09sol

# ps09sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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11/7/2010 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Problem Set 9: Practice Problems Exam 2 Problem 1: Simple Pendulum by Energy Method Solution A simple pendulum consists of a massless string of length l and a point like object of mass m is attached to one end. Suppose the string is fixed at the other end and is initially pulled out at an angle of 0 ! from the vertical and released at rest. a. Use the fact that the energy is constant to find a differential equation describing how the second derivative of the angle ! the object makes with the vertical varies in time. b. Find an expression for the angular velocity of the object at the bottom of its swing. c. Now assume that the initial angle 0 1 rad ! << and thus use the small angle approximation sin ! ! " to show that the simple pendulum behaves like a simple harmonic oscillator. d. Also use the approximation 2 0 0 cos 1 2 ! ! " # , to find an expression for the angular velocity of the object at the bottom of its swing. Solution: We can use energy methods to find the differential equation describing the time evolution of the angle ! . Choose the coordinate system shown in the figure below. When the string is at an angle ! with respect to the vertical, the gravitational potential energy (relative to a choice of zero potential energy at the bottom of the swing where ! = 0 ) is given by

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11/7/2010 2 U = mgl 1 ! cos " ( ) (1.1) The component of the velocity of the object is given by v ! = l ( d ! / dt ) so the kinetic energy is K = 1 2 mv ! 2 = 1 2 m l d ! dt " # \$ % & ' 2 . (1.2) The energy of the system is then E = K + U = 1 2 m l d ! dt " # \$ % & ' 2 + mgl 1 ( cos ! ( ) (1.3) Since there is no non-conservative work, the energy is constant hence 0 = dE dt = 1 2 m 2 l 2 d ! dt d 2 ! dt 2 + mgl sin ! d ! dt = ml 2 d ! dt d 2 ! dt 2 + g l sin ! " # \$ % & ' . (1.4) There are two solutions to this equation, the first one d ! / dt = 0 is the equilibrium solution, the angular speed is zero means the suspended object is not moving. The second solution is the one we are interested in d 2 ! dt 2 + g l sin ! = 0 . (1.5) We can also find the component of the angular velocity at the bottom of the arc using energy methods. When we release the bob from rest, the energy is only potential energy ( ) 0 0 1 cos E U mgl ! = = " (1.6) When the bob is at the bottom of the arc, the only contribution to the energy is the kinetic energy given by 2 2 2 1 1 1 1 1 2 2 d K mv ml dt ! " # = = \$ % & ' . (1.7) Since the energy is constant, we have that U 0 = K 1 or
11/7/2010 3 ( ) 2 2 0 1 1 1 cos 2 d mgl ml dt ! ! " # \$ = % & ' ( (1.8) Hence ( ) 0 1 2 1 cos / d g l dt ! ! " # = ± \$ % & ' ( . (1.9) Using the small angle approximation sin ! ! " , the differential equation Eq. (1.5) becomes 2 2 0 d g dt l ! ! + " , (1.10) which is the simple harmonic oscillator equation. Use 2 0 0 cos 1 2 ! ! " # in Eq. (1.9) to find the component of the angular velocity at the bottom of the arc, ( ) 0 0 1 2 1 cos / d g g l dt l ! ! ! " # = ± \$ = ± % & ' ( , (1.11) in agreement with our result when we used force methods.

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