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1163_Physics ProblemsTechnical Physics

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

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504 Quantum Mechanics *P41.37 (a) For the center of mass to be fixed, m v m v 1 1 2 2 0 + = . Then v v v v m m v m m m v = + = + = + 1 2 1 1 2 1 2 1 2 1 and v m v m m 1 2 1 2 = + . Similarly, v m m v v = + 2 1 2 2 and v m v m m 2 1 1 2 = + . Then 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 1 2 2 2 1 2 1 2 1 2 2 2 2 2 2 m v m v kx m m v m m m m v m m kx m m m m m m v kx v kx + + = + + + + = + + + = + b g b g b g b g µ (b) d dx v kx 1 2 1 2 0 2 2 µ + F H G I K J = because energy is constant 0 1 2 2 1 2 2 = + = + = + µ µ µ v dv dx k x dx dt dv dx kx dv dt kx . Then µ a kx = − , a kx = − µ . 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 a x = − ω 2 , ω µ π = = k f 2 and f k = 1 2 π µ . 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
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