MAT 275 Written Homework #6 SOLUTIONS
3.7, 3.8 Due: March 21
Solve the following problems, showing any necessary work.
1. A spring is hung from the ceiling. When a 20 kg weight is attached to the end of the spring, it stretches by 30 cm. The
spring is stretched another 20 cm, and the weight is let go at this point. (Use g= 9.8 m/s2here.)
a. [1 point] How far is the weight below the natural length of the spring at time t, assuming that γ= 100 kg/s? Describe
the motion as well, using one of the following terms: undamped, underdamped, critically damped, overdamped.
Solution: First, construct the differential equation.
Hooke’s Law states that the force exerted by a spring is F=kd, where dis the distance that the spring has
been stretched. This is equal to the force of gravity F=mg acting on the weight, since the spring and gravity keep
the weight at rest. Hence kd =mg, or k=mg
d=20 ·9.8
0.3=1960
3N/m. The differential equation describing this
system is then given by my00 +γy0+ky = 0; with the intial conditions, this equation is
20y00 + 100y0+1960
3y= 0
y(0) = 0.20
y0(0) = 0
This problem is solved using methods from section 3.1; after multiplication by 3and division by 20, the auxiliary
equation is 3r2+ 15r+ 98 = 0. The solutions are complex numbers: r=−5
2±√951
6i, so the general solution is
y(t) = C1e−5t/2sin √951 t
6!+C2e−5t/2cos √951 t
6!
Its derivative is
y0(t) = −5C1
2−C2√951
6!e−5t/2sin √951 t
6!+ C1√951
6−5C2
2!e−5t/2cos √951 t
6!
The equation y(0) = 0.20 implies C2= 0.20, and y0(0) = 0 implies that
C1√951
6−5C2
2= 0,or C1=15
√951 C2= 0.09728166528.
Thus
y(t) = 0.09728166 e−5t/2sin √951 t
6!+ 0.2e−5t/2cos √951 t
6!or
= 0.097282 e−2.5tsin (5.139715 t)+0.2e−2.5tcos (5.139715 t).
Since the roots of the auxiliary equation are complex (with negative real part), the problem is underdamped.
b. [1 point] Find the value of γfor which the motion is critically damped.
Solution: This is the value γwhich makes the auxiliary equation 20r2+γr +1960
3= 0 have one real root with
multiplicity two. This happens when the discriminant b2−4ac = 0, or when
γ2−4·20 ·1960
3= 0.
This is when γ=r80 ·1960
3≈228.6190427 kg/s.
=⇒
1
c. [2 points] Assume again that γ= 100, and that the weight is also subject to a force whose strength at time tis
F(t) = 8 cos(6t). Find the position of the weight at time t, and find the transient and steady-state solutions. Write
the steady-state solution in the form ys(t) = Rcos(ωt −δ).
Solution: This involves finding the solution to a nonhomogeneous differential equation (with constant coefficients),
which is a four-step process, described in section 3.5.
First, find the solution to the homogeneous version. You found out in part (a) that this is
yh=C1e−5t/2sin √951 t
6!+C2e−5t/2cos √951 t
6!
for some constants C1and C2.
The next step is to determine the form of the solution to the original differential equation
20y00 + 100y0+1960
3y= 8 cos(6t)
y(0) = 0.20
y0(0) = 0
The presence of cos(6t)means that we should expect yp=Acos(6t) + Bsin(6t). Since this is not a solution to the
homogeneous version of the equation, we do not need to multiply by any powers of t.
Step three is to determine Aand B. For this, you need to “substitute” ypinto the differential equation; this
means you need to know that
yp=Acos(6t) + Bsin(6t)
y0
p=−6Asin(6t)+6Bcos(6t)
y00
p=−36Acos(6t)−36Bsin(6t)
Now, pick out the cos(6t)and sin(6t)terms, and make sure that they balance:
20[−36A] + 100[6B] + 1960
3[A]=8 or −200A+ 1800B= 24 [cos(6t)]
20[−36B] + 100[−6A] + 1960
3[B]=0 or −200B−1800A= 0 [sin(6t)]
The second equation implies that B=−9A. Substituting this into the first equation produces A=−3
2050 and
B=27
2050, so
yp=−3
2050 cos(6t) + 27
2050 sin(6t).
Step four combines yhand ypto produce the general solution:
y=C1e−5t/2sin √951 t
6!+C2e−5t/2cos √951 t
6!+−3
2050 cos(6t) + 27
2050 sin(6t)
The initial conditions y(0) = 0.20 and y0(0) = 0 imply that C1=413
2050 and C2=1741√951
649850 , so
y(t) = 413
2050 e−5t/2sin √951 t
6!+1741√951
649850 e−5t/2cos √951 t
6!+−3
2050 cos(6t) + 27
2050 sin(6t)
≈0.082618 e−2.5tsin (5.139715 t)+0.201463 e−2.5tcos (5.139715 t)−0.001463 cos (6t)+0.013171 sin (6t)
=⇒
2
(Part (c) solution, continued.)
The transient solution consists of the terms which look like enegative real number sin(rt)or
enegative real number cos(rt), namely
yt= 0.082618 e−2.5tsin (5.139715 t)+0.201463 e−2.5tcos (5.139715 t).
The steady-state solution consists of the rest of the terms of y(t), namely
ys=−0.001463 cos (6t)+0.013171 sin (6t).
To write the steady-state solution in the form Rcos(ωt −δ), you need to have Rcos δ=−0.001463 and
Rsin δ= 0.013171. This means
R=p(−0.001463)2+ (0.013171)2≈0.013252004,and
t= arctan 0.013171
−0.001463+π≈1.681420216,
(πwas added to the arctangent because cos δ < 0.) Thus
ys=0.013252004 cos(6t−1.681420216).
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