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MTHE 225
HOMEWORK SET 12
1. A block of mass of 100 g stretches a spring 5 cm. Assume there is no damping and
gravitational constant g = 980 cm/sec2 .
(a) If the mass is set in motion from its equilibrium position with a downward velocity
of 10 cm/s, determine the position of the mass at any time.
(b) When does the mass first return to its equilibrium position?
2. A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a
viscous damper with a damping constant of 400 g/s. If the mass is pulled down an
additional 2 cm and then released, find its position at any time.
3. Consider a mass-spring system where mass M = 1, damping constant c = 4, spring
constant k = 4 and external force Fext = 10 cos (3t). Find the formula for the steadystate oscillation, and find its amplitude.
4. A clock designer has a 500 g weight to use in a pendulum clock. Find the length of
pendulum rod that would produce 1 oscillation per one 1 second, that is frequency 1
cycle/sec (a clock maker’s favorite).
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SOLUTION
1.
(a) Given mass M = 100, damping constant c = 0 (since there is damping), and external
force Fext = 0. Now notice that when the mass-spring system is in equilibrium condition
(that is no oscillation), weight of the block is equal to the pulling force of the spring,
that is M g = kL where M is the mass, k is the spring constant and L is the length of
the spring at equilibrium. That means k = M g/L = (100 · 980)/5 = 19600 g/sec2 .
Let y(t) be the displacement of the mass from the equilibrium position at time t. The
differential equation for a mass-spring system is given by
M y 00 + cy 0 + ky = Fext .
Plugging the known values the differential equation becomes
100y 00 + 19600y = 0 ⇒ y 00 + 196 = 0.
√
So the auxiliary equation is: m2 + 196 = 0 ⇒ m2 = −196 ⇒ m = ± −196 = 0 ± 14i.
Hence the general solution is
y(t) = e0·t [C1 cos (14t) + C2 sin (14t)] ⇒ y(t) = C1 cos (14t) + C2 sin (14t)
Now from the question we see that y(0) = 0 and y 0 (0) = 10.
We see that, y 0 (t) = −14C1 sin (14t) + 14C2 cos (14t)
y(0) = 0 ⇒ 0 = C1 cos 0 + C2 sin 0 ⇒ C1 = 0.
y 0 (0) = 10 ⇒ 10 = −14C1 sin 0 + 14C2 cos 0 ⇒ C2 = 10/14
10
So, y(t) =
sin (14t).
14
10
sin (14t) ⇒
14
sin (14t) = 0 ⇒ 14t = πn for some integer n. Since n = 0 means the starting point, we
π
should use n = 1 in order to get our required answer, that is t =
sec.
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(b) The mass returns to equilibrium position means y(t) = 0. That means 0 =
2.
Given mass M = 20, damping constant c = 400 g/sec, and external force Fext = 0. Now
notice that when the mass-spring system is in equilibrium condition (that is no oscillation),
weight of the block is equal to the pulling force of the spring, that is M g = kL where M
is the mass, k is the spring constant and L is the length of the spring at equilibrium. That
means k = M g/L = (20 · 980)/5 = 3920 g/sec2 .
2
Let y(t) be the displacement of the mass from the equilibrium position at time t. The
differential equation for a mass-spring system is given by
M y 00 + cy 0 + ky = Fext .
Plugging the known values the differential equation becomes
20y 00 + 400y 0 + 3920y = 0 ⇒ y 00 + 20y 0 + 196 = 0.
−20 ±
So the auxiliary equation is: m2 + 20m + 196 = 0 ⇒ m =
√
m = −10 ± 4 6i. Hence the general solution is
√
√
y(t) = e−10t [C1 cos (4 6t) + C2 sin (4 6t)].
√
202 − 4 · 1 · 196
⇒
2
Now from the question we see that y(0) = 2 and y 0 (0) = 0.
We see that,
√
√
√
√
√
y 0 (t) = −10e−10t [C1 cos (4√ 6t)+C2 sin√(4 6t)]+4√ 6e−10t [−C1 sin (4 √6t)+C2 cos (4 6t)]
= e−10t [(−10C1 + 4 6C2 ) cos (4 6t) + (−4 6C1 − 10C2 ) sin (4 6t)]
y(0) = 2 ⇒ 2 = C1 cos 0 + C2 sin 0 ⇒ C1 = 2.
√
√
√
)
sin
0+(−10C
+4
6C
)
cos
0
⇒
(−10C
+4
6C2 ) = 0
y 0 (0) = 0 ⇒ 0 =√(−4 6C1 −10C
2
1
2
1
√
√
⇒ C2 = 10C1 /4 6 = 20/4 6 = 5/ 6
√
√
5
−10t
So, y(t) = e
2 cos (4 6t) + √ sin (4 6t) .
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3.
The differential equation for any mass-spring system is
M y 00 + cy 0 + ky = Fext .
Plugging the known values this differential equation becomes
y 00 + 4y 0 + 4 = 10 cos (3t)
• For yc : Set y 00 + 4y 0 + 4 = 0.
So the auxiliary equation is m2 + 4m + 4 = 0 ⇒ (m + 2)2 = 0 ⇒ m = −2, −2.
Therefore, yc (t) = C1 e−2t + C2 te−2t
Recall yc (t) is known as “transient part” of the general solution since yc (t) → 0 as
t → ∞.
• For yp (t) : Try yp (t) = A cos (3t) + B sin (3t). So we have
yp0 (t) = −3A sin (3t) + 3B cos (3t) and yp00 (t) = −9A cos (3t) − 9B sin (3t)
Now plugging these back in the main differential equation we get
−9A cos (3t) − 9B sin (3t) + 4(−3A sin (3t) + 3B cos (3t)) + 4(A cos (3t) + B sin (3t)) =
10 cos (3t)
3
⇒ (−9A + 12B + 4A) cos (3t) + (−9B − 12A + 4B) sin (3t) = 10 cos (3t)
⇒ (−5A + 12B) cos (3t) + (12A − 5B) sin (3t) = 10 cos (3t)
Now comparing the coefficient we get: −5A + 12B = 10 and −12A − 5B = 0,
that is: −60A + 144B = 120 and −60A − 25B = 0,
that is: (−60A + 144B) − (60A − 25B) = 120 − 0,
that is: 169B = 60 ⇒ B = 120/169.
5B
5 120
50
120
we have A = −
=− ·
=−
.
Now since −12A − 5B = 0 and B =
169
12
12 169
169
50
120
So yp (t) = −
cos (3t) +
sin (3t)
169
169
Recall yp (t) is known as “steady state part” of the general solution.
50
120
Now choose two constants C and θ such a that C cos θ = −
and C sin θ =
.
169
169
s
2 2
120
10
50
+
=
That means C =
−
169
169
13
10
10
Now plugging them in yp (t) we get yp (t) =
cos (3t − θ). So the amplitude is
.
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13
4.
Recall the motion of pendulum is described by
θ00 +
g
sin θ = 0
L
where θ is the angular position/displacement of the pendulum, g is the gravitational acceleration and L is the length of the pendulum rod.
Now by choose θ < 10◦ we can use the approximation sin θ ≈ θ and the differential
equation becomes:
g
θ00 + θ = 0
L
r
r
g
g
g
2
So the auxiliary equation is: m + = 0 ⇒ m = ± − = 0 ±
i
L
L
L
Therefore the general solution is:
r
r
r
r
g
g
g
g
0·t
y(t) = e
C1 cos
t + C2 sin
t
= C1 cos
t + C2 sin
t
L
L
L
L
Now choose two constants φ and A such that C1 = A cos φ and C2 = A sin φ.
r
g
Now plugging these in y(t) we get y(t) = A cos
t − φ = A cos (bt − φ) where
L
r
g
b=
.
L
4
s
2π
L
So the period of oscillation is:
=
2π.
b
g
r
1
g 1
So the frequency is:
=
period
L 2π
r
g 1
g 1
g
So we want:
=1⇒
=
1
⇒
L
=
L 2π
L 4π 2
4π 2
Remark: Notice neither y(t) nor period nor frequency does not depend up on the mass of
the pendulum.
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