problem13_23

University Physics with Modern Physics with Mastering Physics (11th Edition)

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13.23: a) Setting 2 2 1 2 2 1 kx mv = in Eq. (13.21) and solving for x gives . 2 A x ± = Eliminating x in favor of v with the same relation gives . 2 2 2 ωA x m kA v ± = ± = b) This happens four times each cycle, corresponding the four possible combinations of + and –
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Unformatted text preview: in the results of part (a). The time between the occurrences is one-fourth of a period or ( 29 8 3 8 4 3 4 1 2 4 2 2 2 , , c) . 4 / kA kA ω π ω π K U E K E U T = = = = = =...
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  • kx, possible combinations, Ford Ka

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