105Sol6 - Physics 105 Solutions to Problem Set 6 By Stephen...

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Physics 105 Solutions to Problem Set 6 By Stephen Glassman, Fall 2007 8.22 If F = kr - 3 , then U = 1 2 kr - 2 . We substitute this potential into equation (8.30). U eff ( r ) = U ( r ) + ` 2 2 μr 2 = 1 r 2 ± 1 2 k + ` 2 2 μ ² So U eff ( r ) = αr - 2 where α = 1 2 ( k + ` 2 μ - 1 ) can be positive, negative, or zero depend- ing on k . Such graphs are simple enough that I won’t include them here. For α = 0, every r is a critical point, and r satisfies ¨ r = 0. For α positive, there are no critical points. The potential is infinite at the origin, so every orbit heads to infinity. For α negative, again there are no critical points. The potential at infinity is zero, so whether the orbit is bounded depends on whether the effective 1D motion has positive energy. (b) Since u = r - 1 , F = ku 3 , and equation (8.41) becomes: u 00 ( φ ) = - u ( φ ) - μ ` 2 u ( φ ) 2 F = - u ( φ ) - ` 2 u ( φ ) = - ± 1 + ` 2 ² u ( φ ) = - μ ` 2 αu ( φ ) For α = 0, the solution is u ( φ ) = A + . So r goes to infinity at some (possibly negative) φ . Also, r goes to 0 at large positive and negative φ . For α > 0, this is the usual differential equation for an harmonic oscillator (in φ not t ). Its solutions are u ( φ ) = A cos (p 1 + μ ` 2 φ + δ ) . Note that u has zeros but no infinities, so r goes to infinity but not zero just as predicted. For α < 0, the solutions are exponential growth and decay. u ( φ ) = Ae βφ + Be - βφ where β = p - μ ` 2 α . This function only has zeros if A and B have opposite signs. Therefore, r definitely has zeroes and might reach infinity. 1
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8.23 (a) If F = - ku 2 + λu 3 , then equation (8.41) becomes: u 00 ( φ ) = - u ( φ ) - m ` 2 u ( φ ) 2 F = - u ( φ ) - λm ` 2 u ( φ ) + km ` 2 = - ± 1 + λm ` 2 ² u ( φ ) + km ` 2 This is the same differential equation we encountered in our discussion of a mass on a spring in gravity. (See midterm solutions.) The homogenous solutions are the harmonic oscillator solutions, plus there is a φ independent particular solution. Our most general solution is then: u ( φ ) = A cos( r 1 + λm ` 2 φ + δ ) + km ` 2 ± 1 + λm ` 2 ² - 1 = A cos( r 1 + λm ` 2 φ + δ ) + ± ` 2 km + λ k ² - 1 Now in the manner of § 8.6, we’ll choose coordinates so that δ = 0.
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105Sol6 - Physics 105 Solutions to Problem Set 6 By Stephen...

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