43(4.0 10 N/m)(2.0 10 m)=80 N.Fkx−== ××(e) At half of the maximum displacement, 1.0 mmx=, 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 am= ω2xm, where is the angular frequency and xm= 0.0020 m is the amplitude. Thus, am= 8000 m/s2leads to = 2000 rad/s. Using Newton’s second law with m= 0.010 kg, we have ==+=8020003Fmamattm−−−FHIKcosN cosωφafchafπwheretis understood to be in seconds. (a) Eq. 15-5 gives T= 2π/= 3.1 × 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.