, c
Solutions Manual ◯
to accompany
System Dynamics, Third Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter One
◯c Solutions Manual Copyright 2013 The McGraw-Hill Companies. All rights
reserved. No part of this manual may be displayed, reproduced, or distributed
in any form or by any means without the written permission of the publisher
or used beyond the limited distribution to teachers or educators permitted by
McGraw-Hill for their individual course preparation. Any other reproduction
or translation of this work is unlawful.
,1.1 W = mg = 3(32.2) = 96.6 lb.
1.2 m = W/g = 100/9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) =
0.698 slug.
1.3 d = (50 + 5/12)(0.3048) = 15.37 m.
1.4 d = 3(100)(0.3048) = 91.44 m
1.5 d = 100(3.281) = 328.1 ft
1.6 d = 50(3600)/5280 = 34.0909 mph
1.7 v = 100(0.6214) = 62.14 mph
1.8 n = 1/[60(1.341 × 10−3)] = 12.43, or approximately 12 bulbs.
1.9 5(70 − 32)/9 = 21.1◦ C
1.10 9(30)/5 + 32 = 86◦ F
1.11 ω = 3000(2π)/60 = 314.16 rad/sec. Period P = 2π/ω = 60/3000 = 1/50 sec.
1.12 ω = 5 rad/sec. Period P = 2π/ω = 2π/5 = 1.257 sec. Frequency f = 1/P = 5/2π =
0.796 Hz.
1.13 Speed = 40(5280)/3600 = 58.6667 ft/sec. Frequency = 58.6667/30 = 1.9556 times
per second.
1.14 x = 0.005 sin 6t, x˙ = 0.005(6) cos 6t = 0.03 cos 6t. Velocity amplitude is 0.03 m/s.
ẍ = −6(0.03) sin 6t = −0.18 sin 6t. Acceleration amplitude is 0.18 m/s2. Displacement,
velocity and acceleration all have the same frequency.
1.15 Physical considerations require the model to pass through the origin, so we seek a
model of the form f = kx. A plot of the data shows that a good line drawn by eye is given
by f = 0.2x. So we estimate k to be 0.2 lb/in.
◯c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
, 1.16 The script file is
x = [0:0.01:1];
subplot(2,2,1)
plot(x,sin(x),x,x),xlabel('x (radians)'),ylabel('x and sin(x)'),...
gtext('x'),gtext('sin(x)')
subplot(2,2,2)
plot(x,sin(x)-x),xlabel('x (radians)'),ylabel('Error: sin(x) - x')
subplot(2,2,3)
plot(x,100*(sin(x)-x)./sin(x)),xlabel('x (radians)'),...
ylabel('Percent Error'),grid
The plots are shown in the figure.
1 0
0.8
−0.05
Error: sin(x) − x
x and sin(x)
0.6
−0.1
0.4
−0.15
0.2
0 −0.2
0 0.5 1 0 0.5 1
x (radians) x (radians)
0
−5
Percent Error
−10
−15
−20
0 0.5 1
x (radians)
Figure : for Problem 1.16.
From the third plot we can see that the approximation sin x ≈ x is accurate to within
5% if |x| ≤ 0.5 radians.
◯c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Solutions Manual ◯
to accompany
System Dynamics, Third Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter One
◯c Solutions Manual Copyright 2013 The McGraw-Hill Companies. All rights
reserved. No part of this manual may be displayed, reproduced, or distributed
in any form or by any means without the written permission of the publisher
or used beyond the limited distribution to teachers or educators permitted by
McGraw-Hill for their individual course preparation. Any other reproduction
or translation of this work is unlawful.
,1.1 W = mg = 3(32.2) = 96.6 lb.
1.2 m = W/g = 100/9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) =
0.698 slug.
1.3 d = (50 + 5/12)(0.3048) = 15.37 m.
1.4 d = 3(100)(0.3048) = 91.44 m
1.5 d = 100(3.281) = 328.1 ft
1.6 d = 50(3600)/5280 = 34.0909 mph
1.7 v = 100(0.6214) = 62.14 mph
1.8 n = 1/[60(1.341 × 10−3)] = 12.43, or approximately 12 bulbs.
1.9 5(70 − 32)/9 = 21.1◦ C
1.10 9(30)/5 + 32 = 86◦ F
1.11 ω = 3000(2π)/60 = 314.16 rad/sec. Period P = 2π/ω = 60/3000 = 1/50 sec.
1.12 ω = 5 rad/sec. Period P = 2π/ω = 2π/5 = 1.257 sec. Frequency f = 1/P = 5/2π =
0.796 Hz.
1.13 Speed = 40(5280)/3600 = 58.6667 ft/sec. Frequency = 58.6667/30 = 1.9556 times
per second.
1.14 x = 0.005 sin 6t, x˙ = 0.005(6) cos 6t = 0.03 cos 6t. Velocity amplitude is 0.03 m/s.
ẍ = −6(0.03) sin 6t = −0.18 sin 6t. Acceleration amplitude is 0.18 m/s2. Displacement,
velocity and acceleration all have the same frequency.
1.15 Physical considerations require the model to pass through the origin, so we seek a
model of the form f = kx. A plot of the data shows that a good line drawn by eye is given
by f = 0.2x. So we estimate k to be 0.2 lb/in.
◯c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
, 1.16 The script file is
x = [0:0.01:1];
subplot(2,2,1)
plot(x,sin(x),x,x),xlabel('x (radians)'),ylabel('x and sin(x)'),...
gtext('x'),gtext('sin(x)')
subplot(2,2,2)
plot(x,sin(x)-x),xlabel('x (radians)'),ylabel('Error: sin(x) - x')
subplot(2,2,3)
plot(x,100*(sin(x)-x)./sin(x)),xlabel('x (radians)'),...
ylabel('Percent Error'),grid
The plots are shown in the figure.
1 0
0.8
−0.05
Error: sin(x) − x
x and sin(x)
0.6
−0.1
0.4
−0.15
0.2
0 −0.2
0 0.5 1 0 0.5 1
x (radians) x (radians)
0
−5
Percent Error
−10
−15
−20
0 0.5 1
x (radians)
Figure : for Problem 1.16.
From the third plot we can see that the approximation sin x ≈ x is accurate to within
5% if |x| ≤ 0.5 radians.
◯c 2013 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
