Ch15

103. Using Eq. 15-12, we find ω = k / m = 10 rad / s . We also use vm = xmω and am = xmω2. (a) The amplitude (meaning “displacement amplitude”) is xm = vm/ω = 3/10 = 0.30 m. (b) The acceleration-amplitude is am = (0.30)(10)2 = 30 m/s2. (c) One interpretation of this question is “what is the most negative value of the acceleration?” in which case the answer is –am = –30 m/s2. Another interpretation is “what is the smallest value of the absolute-value of the acceleration?” in which case the answer is zero. (d) Since the period is T = 2π/ω = 0.628 s. Therefore, seven cycles of the motion requires t = 7T = 4.4 s.