In this problem we derive a formula for the natural period of an undamped nonlinear

Chapter 9, Problem 26

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In this problem we derive a formula for the natural period of an undamped nonlinear pendulum (equation (10) with c = 0 in Section 9.2). Suppose that the bob is pulled through a positive angle and then released with zero velocity. a. We usually think of and d/dt as functions of t. However, if we reverse the roles of t and , we can regard t as a function of and, consequently, can also think of d/dt as a function of . Then derive the following sequence of equations: 1 2 mL2 d d __d dt _2_ = mgL sin , 1 2 m_L d dt _2 = mgL(cos cos ), dt = _ L 2g d _cos cos . Why was the negative square root chosen in the last equation? b. If T is the natural period of oscillation, derive the formula T 4 = _ L 2g _ 0 d _cos cos . c. Use the identities cos = 1 2 sin2_ 2 _ and cos = 1 2 sin2_2 _, followed by the change of variable sin_ 2 _ = k sin with k = sin_2 _, show that T = 4_L g _ /2 0 d _1 k2 sin2 . The integral is called the elliptic integral of the first kind. Note that the period depends on the ratio L/g and also on the initial displacement through k = sin(/2). N d. By evaluating the integral in the expression for T , obtain values for T that you can compare with the graphical estimates you obtained in 20. 2