CLICK HERE ACCESS THEThisCOMPLETE Solutions © 2016 Pearson Education, Inc., Upper Saddle River,TO NJ. All rights reserved. material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 22–1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.
SOLUTION $ mg - k(y + yst) = my
+ T ΣFy = may;
where kyst = mg
k $ y + y = 0 m
Hence
p =
=
k Bm
B
Where k =
8(9.81) 0.175
448.46 = 7.487 8 $ y + (7.487)2y = 0
6
= 448.46 N>m
$ y + 56.1y = 0
Ans.
The solution of the above differential equation is of the form: y = A sin pt + B cos pt
(1)
# v = y = Ap cos pt - Bp sin pt
(2)
At t = 0, y = 0.1 m and v = v0 = 1.50 m>s From Eq. (1)
0.1 = A sin 0 + B cos 0
From Eq. (2)
v0 = Ap cos 0 - 0
Hence
y = 0.2003 sin 7.487t + 0.1 cos 7.487t
At t = 0.22 s,
y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]
B = 0.1 m A =
v0 1.50 = = 0.2003 m p 7.487
Ans.
= 0.192 m
1190
Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m