System dynamics william palm III 1ed ISM

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Solutions Manual c

to accompany System Dynamics, First Edition

by William J Palm III University of Rhode Island

Solutions to Problems in Chapter One

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c

1.6 ω = 3000(2π)/60 = 314.16 rad/sec Period P = 2π/ω = 60/3000 − 1/50 sec.

1.7 ω = 5 rad/sec Period P = 2π/ω = 2π/5 = 1.257 sec Frequency f = 1/P = 5/2π =

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4.912 3.293 =

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y(t) = 3(0.094) cos(3t + 0.559) = 0.282 cos(3t + 0.559)

The amplitude of the velocity is 0.282 ft/sec

For the acceleration,

¨

y(t) = −3(0.282) sin(3t + 0.559) = 0.846 sin(3t + 0.559)

The amplitude of the acceleration is 0.846 ft/sec2

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1.16 ˙y = 5(0.07) sin(5t + φ) Thus ˙ y(0) = 0.35 sin φ = 0.041, and

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1.18 The radian frequency is 2π(10) = 20π The displacement is y = A sin 20πt The

velocity and acceleration are

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1.20 y certainly lies within the range [−6, 6] and y disappears after about four time

con-stants, 4(3) = 12

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1.22 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

1.23 The script file is

x = [0:0.01:1];

subplot(2,2,1)

plot(x,sin(x),x,x),xlabel(0x (radians)0),ylabel(0x and sin(x)0),

gtext(0x0),gtext(0sin(x)0)

subplot(2,2,2)

plot(x,sin(x)-x),xlabel(0x (radians)0),ylabel(0Error: sin(x) - x0)

subplot(2,2,3)

plot(x,100*(sin(x)-x)./sin(x)),xlabel(0x (radians)0),

ylabel(0Percent Error0),grid

The plots are shown in the figure

Figure : for Problem 1.23

From the third plot we can see that the approximation sin x ≈ x is accurate to within 5% if |x| ≤ 0.5 radians.

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θ − 3π

4



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1.26 For h near 25,

f (h) ≈

√

25 + 12

√

25(h − 25) = 5 +

1

10(h − 25)

1.27 For r near 5,

f (r) ≈ 52+ 2(5)(r − 5) = 25 + 10(r − 5) For r near 10,

f (r) ≈ 102+ 2(10)(r − 10) = 100 + 20(r − 10)

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1.28 For h near 16,

f (h) ≈

√

16 + 12

1.29 Construct a straight line the passes through the two endpoints at p = 0 and p = 900.

At p = 0, f (0) = 0 At p = 900, f (900) = 0.002√900 = 0.06 This straight line is

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1.30 (a) The data is described approximately by the linear function y = 54x − 1360 The

precise values given by the least squares method are y = 53.5x − 1354.5 (see Problem 1.48a).

(b) Only the loglog plot of the data gives something close to a straight line, so the data

is best described by a power function y = bxm where the approximate values are m = −0.98 and b = 3600 The precise values given by the least squares method are y = 3582.1x−0.9764

(see Problem 1.48b)

(c) Both the loglog and semilog plot (with the y axis logarithmic) give something close

to a straight line, but the semilog plot gives the straightest line, so the data is best described

by a exponential function y = b(10)mx where the approximate values are m = −0.007 and

b = 2.1 × 105 The precise values given by the least squares method are y = 2.0622 ×

105(10)−0.0067x (see Problem 1.48c)

1.31 With this problem, it is best to scale the data by letting x = year − 1990, to avoid

raising large numbers like 1990 to a power Both the loglog and semilog plot (with the

y axis logarithmic) give something close to a straight line, but the semilog plot gives the

straightest line, so the data is best described by a exponential function y = b(10)mx The

approximate values are m = 0.035 and b = 9.98.

Set y = 20 to determine how long it will take for the population to increase from 10 to

20 million This gives 20 = 9.98(10)0.03x Solve it for x: x = (log(20) − log(9.98))/0.035.

The answer is 8.63 years, which corresponds to 8.63 years after 1990

More precise values are given by the least squares method (see Problem 1.49)

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1.32 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e , which gives b =

− ln(0.5)/5500 = 1.2603 × 10−4

(b) Solve for t to obtain t = − ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603 ×

10−4 The answer is 836 years Thus the organism died 836 years ago

(c) Using b = 1.1(1.2603 × 10−4) in t = − ln(0.9)/b gives 760 years Using b = 0.9(1.2603 × 10−4) in t = − ln(0.9)/b gives 928 years.

1.33 Only the semilog plot of the data gives something close to a straight line, so the

data is best described by an exponential function y = b(10)mx where y is the temperature

in degrees C and x is the time in seconds The approximate values are m = −3.67 and

b = 356 The alternate exponential form is y = be(m ln 10)x = 356e−8.451x The time

constant is 1/8.451 = 0.1183 s.

The precise values given by the least squares method are y = 356.0199(10)−3.6709x (seeProblem 1.51)

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1.34 Only the semilog plot of the data gives something close to a straight line, so the data is

best described by an exponential function y = b(10)mx where y is the bearing life thousands

of hours and x is the temperature in degrees F The approximate values are m = −0.007 and b = 142 The bearing life at 150 ◦ F is estimated to be y = 142(10)−0.007(150) = 12.66,

or 12,600 hours The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x The

time constant is 1/0.0161 = 62.1 or 6.21 × 104 hr

The precise values given by the least squares method are y = 141.8603(10)−0.0070x (seeProblem 1.52)

1.35 Only the semilog plot of the data gives something close to a straight line, so the

data is best described by an exponential function y = b(10)mx where y is the voltage and

x is the time in seconds The first data point does not lie close to the straight line on

the semilog plot, but a measurement error of ±1 volt would account for the discrepancy

The approximate values are m = −0.43 and b = 96 The alternate exponential form is

y = be(m ln 10)x= 96e−0.99x The time constant is 1/0.99 = 1.01 s.

The precise values given by the least squares method are y = 95.8063(10)−0.4333x (seeProblem 1.53)

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1.36 A semilog plot generated by the following script file shows that the exponential function

T − 70 = bemt fits the data well

1.37 In the first printing of the text, the time data was reversed (the largest height should

have the smallest time) This misprint also occurs in the data table on page 28 Plots ofthe data on a log-log plot and rectilinear scales both give something close to a straight line,

so we try both functions (Note that the flow should be 0 when the height is 0, so we donot consider the exponential function and we must force the linear function to pass through

the origin by setting b = 0.) The three lowest heights give the same time, so we discard the

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1.38 In the first printing of the text, the time data was reversed (the largest height should

have the smallest time) This misprint also occurs in the data table on page 28 Plots ofthe data on a log-log plot and rectilinear scales both give something close to a straight line,

so we try both functions (Note that the flow should be 0 when the height is 0, so we donot consider the exponential function and we must force the linear function to pass through

the origin by setting b = 0.) The variable x is the height and the variable y is the flow rate.

The three lowest heights give the same time, so we discard the heights of 1 and 2 cm

The power function fitted by eye in terms of the height h is approximately f = 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow This is because the

flow through the outlet is pipe flow For the linear function f = mh, the best fit by eye is approximately f = 3.7h.

Using the least squares method gives more precise results: f = 4.1796h0.9381 and f = 3.6735h (see Problem 1.56).

1.39 Fitting a straight line by eye gives the approximate values m = 15 and b = 7 The

precise values given by the least squares method are m = 15.0750 and b = 7.1500 (see

Problem 1.57)

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1.40 Plot the data on a loglog plot We must delete the first data point to avoid taking

the logarithm of 0 The power function fitted by eye is approximately y = 7x3 The precise

values given by the least squares method are m = 2.9448 and b = 7.4053 (see Problem

1.58)

1.41 Plot the data on a semilog plot The exponential function fitted by eye is approximately

y = 6e3x The precise values given by the least squares method are y = 6.4224e2.8984x (seeProblem 1.59)

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1.42 Fitting a straight line by eye through the origin gives the approximate values m = 17

and b = 0 The precise value given by the least squares method is m = 16.6071 (see Problem

1.60)

1.43 Plot the data on a loglog plot We must delete the first data point to avoid taking

the logarithm of 0 The power function fitted by eye is approximately y = 7x3 The precise

value given by the least squares method is b = 7.4793 (see Problem 1.61).

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1.44 a) The equation for b is obtained as follows.

mately y = 6e3x The precise values given by the least squares method are y = 5.8449e3x

(see Problem 1.62)

1.45 The integral form of the sum of squares to fit the function y = 5x is

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1.46 a) The integral form of the sum of squares to fit the function y = Ax + Bx is

J =

Z L 0

b) With L = 2, A = 3, and B = 5, we obtain m = 11 and b = −2 The fitted straight line is y = 11x − 2.

1.47 a) The integral form of the sum of squares to fit the function y = Be is

J =

Z L 0

which can be solved for m and b, given values of M , B, and L.

b) With L = 1, M = −5, and B = 15, we obtain m = −10.973 and b = 8.466 The fitted straight line is y = −10.973x + 8.466.

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1.48 (a) The script file is