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PROBLEM 19.1 Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 5 mm and a period of 0.1 s. SOLUTION Simple harmonic motion. where x = xm sin(w nt + f ) wn = 2p 2p = = 62.83 rad/s T 0.1 s xm = 5 mm = 5 ¥ 10 -3 m Then x = 5 ¥ 10 -3 sin(62.83t + f ) Differentiate to obtain velocity and acceleration. v = x& = w n xm cos(w nt + f ) a = && x = - w n2 xm sin(w nt + f ) Maximum velocity. Maximum acceleration. | x& max | = w n xm = (62.83)(5 ¥ 10 -3 ) | && xmax | = w n2 xm = (62.83)2 (5 ¥ 10 -3 ) x& max = 0.314 m/s && xmax = amax = 19.74 m/s2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3
