... .23cos,43 ωtπAty e) See Exercise s.m 315. 0 ;12 .15 ωA f) From the result of part s.m 315. 0 mm.0,d yvy 15. 14: Solving Eq. (15. 13) for the force ,F .2.43))m750.0()Hz0.40((m2.50kg120.0222fμμvF 15. 15: ... direction of the wave motion. 15. 2: vf Hz105.1m001.0sm15006vf 15. 3: a)m.0.439Hz)(784s)m344( fv b) Hz.105.25m)10(6.55s)m344(65vf 15. 4: Denoting the speed of ... m.87.2b).m556Hz105.104sm1000.3Hz10540sm100036838. 15. 5: a) Hz)000,20(s)m344(m,17.2Hz)0.20()sm344(minmaxcm.72.1 b) mm.74.0Hz)000,20(s)m1480(m,74.0Hz)0.20()sm1480(minmax 15. 6: Comparison with Eq. (15. 4) gives...