ch15-p033 - 33. The textbook notes (in the discussion...

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43 (4.0 10 N/m)(2.0 10 m)=80 N. Fk x == × × (e) At half of the maximum displacement, 1.0 mm x = , and the force is (4.0 10 N/m)(1.0 10 m)=40 N. x × 33. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is a m = ω 2 x m , where is the angular frequency and x m = 0.0020 m is the amplitude. Thus, a m = 8000 m/s 2 leads to = 2000 rad/s. Using Newton’s second law with m = 0.010 kg, we have = = + = 80 2000 3 Fm ama t t m −− F H I K cos N cos ωφ a f ch a f π where t is understood to be in seconds. (a) Eq. 15-5 gives T = 2 π / = 3.1 × 10
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