1163_Physics ProblemsTechnical Physics

1163_Physics ProblemsTechnical Physics - 504 *P41.37...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
504 Quantum Mechanics *P41.37 (a) For the center of mass to be fixed, mv 11 22 0 += . Then vv v v m m v mm m v =+=+ = + 121 1 2 1 21 2 1 and v 1 2 12 = + . Similarly, v m m vv =+ 2 1 and v 2 1 = + . Then 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 12 1 2 2 kx mmv kx mm m m vk x x ++ = + + + + = + + +=+ bg µ (b) d dx x 1 2 1 2 0 + F H G I K J = because energy is constant 0 1 2 2 1 2 2 = + = + µµ v dv dx dx dt dv dx kx dv dt kx . Then ak x =− , a kx . This is the condition for simple harmonic motion, that the acceleration of the equivalent particle be a negative constant times the excursion from equilibrium. By identification with ax ω 2 , π == k f 2 and f k = 1 2 πµ . P41.38 (a) With x = 0 and p x = 0, the average value of x 2 is x a f 2 and the average value of p x 2 is p x 2 . Then x p x = 2
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online