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Unformatted text preview: February 27, 2010 Cornell University, Department of Physics P3318, Analytical Mechanics, HW#4, due: 2/26/10, 2:30PM (at section) Question 1 : A symmetric well potential Consider a particle of mass m in the following one dimensional potential V ( x ) = ax 2 n , (1) where n is a positive integer and a > 0. 1. What are the units of a ? Answer: mx 2(1- n ) t- 2 2. Calculate the period of the particle as a function of its energy. You will need the following integral integraldisplay 1- 1 dx parenleftBig 1 x 2 n parenrightBig- 1 / 2 = 2 parenleftbigg 1 + 1 2 n parenrightbigg parenleftbigg n + 1 2 n parenrightbigg- 1 . (2) Answer: We use the general formula for the period T = integraldisplay B A dx radicalBigg 2 m E V ( x ) (3) In our case the turning points, A and B , are given by B = A = parenleftbigg E a parenrightbigg 1 / (2 n ) (4) We then rescale x such that z = x/B and thus T = integraldisplay 1- 1 Bdz radicalBigg 2 m E aB 2 n z 2 n = E (1- n ) / (2 n ) a 1 /...
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This note was uploaded on 03/29/2010 for the course PHYS 3318 taught by Professor Flanagan during the Spring '08 term at Cornell University (Engineering School).
- Spring '08