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3.7 Spring Systems
253
The resulting amplification of vibration eventually becomes large enough to
destroy the mechanical system. This is a manifestation of resonance discussed further in Section ??.
Exercises
Assume forces are in pounds or Newtons and lengths are in feet or meters.
1–4. Assume Hooke’s Law.
1. A body weighing 16 lbs stretches a spring 6 inches. Find the spring constant.
Solution: The force is 16 lbs. A length of 6 inches is 1/2 ft. The spring
constant is 𝑘 = 16/(1/2) = 32 lbs/ft.
Answer: 𝑘 = 32 lbs/ft
2. The spring constant of a certain spring is 𝑘 = 20 lbs/ft. If a body stretches
the spring 9 inches how much does it weigh?
Solution: The equation for the spring constant gives 20 =
240/9 = 80/3 lbs.
𝑥
9/12 .
So 𝑥 =
Answer: 80/3 lbs.
3. A mass of 40 kilograms stretches a spring 80 centimeters. Find the spring
constant.
Solution: The force exerted by the mass is 40 ⋅ 9.8 = 392 N. Thus 𝑘 =
392/.8 = 490 N/m.
Answer: 𝑘 = 490 N/m
4. The spring constant of a certain spring is 𝑘 = 784 N/m. How far will a
mass of 20 kilograms stretch the spring?
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3 Second Order Constant Coefficient Linear Differential Equations
Solution: The force exerted by the mass is 20 ⋅ 9.8 = 196 N. Let 𝑥 be the
distance stretched. Thus 784 = 196
𝑥 . Solving for 𝑥 gives 𝑥 = 196/784 =
1/4 m or 250 centimeters.
Answer: 250 centimeters
5–8. Assume the damping force of a dashpot is proportional to the velocity
of the body.
5. A dashpot exerts a damping force of 4 lbs when the velocity of the mass
is 6 inches per second. Find the damping constant.
Solution: The force is 4 lbs and the velocity is 1/2 ft per second. So
4
𝜇 = Force/velocity = 1/2
= 8 lbs s/ft.
Answer: 𝜇 = 8 lbs s/ft
6. A dashpot exerts a damping force of 40 Newtons when the velocity is 30
centimeters per second. Find the damping constant.
Solution: The force is 40 Newtons and the velocity is 30 centimeters per
second which is .3 m per second. So 𝜇 = 40/.3 = 133.33 N s/m.
Answer: 𝜇 = 133.33 N s/m
7. A dashpot has a damping constant 𝜇 = 100 lbs s/ft and decelerates a
body by 4 ft per second. What was the force exerted by the body?
Solution: Let 𝑥 be the force. Then 100 = 𝑥/4 so 𝑥 = 400 lbs.
Answer: 400 lbs
8. A force of 40 N is applied on a body connected to a dashpot having
a damping constant 𝜇 = 200 N s/m. By how much will the dashpot
decelerates the body?
Solution: Let 𝑣 be the velocity of deceleration. Then 200 = 40/𝑣 and so
𝑣 = 1/5 m/s or 20 centimeters per second.
Answer: 20 centimeters per second
9–14. Find a mathematical model that represents the motion of the body
and determine the resulting motion.
9. A spring is stretched 10 centimeters by a force of 2 Newtons. A body
of mass 6 kilogram is attached to such a spring with an accompanying
dashpot. At 𝑡 = 0 the mass is pulled down from its equilibrium position
a distance of 10 centimeters and released. Suppose there is no damping.
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Solution: The mass is 𝑚 = 6. The spring constant is given by 𝑘 = 2/.1 =
20. The damping constant is 𝜇 = 0. Since no external force is mentioned
we may assume it is zero. The initial conditions are 𝑦(0) = .1 m and
𝑦 ′ (0) = 0. The following equation
6𝑦 ′′ (𝑡) + 20𝑦 = 0,
𝑦(0) = .1,
𝑦 ′ (0) = 0
represents the model for the motion of the body. The motion is undamped
free or simple harmonic motion.
Answer: 6𝑦 ′′ (𝑡) + 20𝑦 = 0,
simple harmonic motion
𝑦(0) = .1,
𝑦 ′ (0) = 0, undamped free or
10. A body of mass 4 kg will stretch a spring 80 centimeters. This same body
is attached to such a spring with an accompanying dashpot. Suppose the
damping force is 98 N when the velocity of the body is 2 m/s . At 𝑡 = 0
the mass is pulled down from its equilibrium position a distance of 50
centimeters and released with an upward velocity of 10 centimeters per
second.
Solution: The mass is 𝑚 = 4. The spring constant 𝑘 is given by 𝑘 =
4 ⋅ 9.8/.8 = 49. The damping constant is given by 𝜇 = 98/2 = 49. Since
no external force is mentioned we may assume it is zero. The initial
conditions are 𝑦(0) = .5 and 𝑦 ′ (0) = −.1. The equation 4𝑦 ′′ (𝑡) + 49𝑦 ′(𝑡) +
49𝑦 = 0, 𝑦(0) = .5, 𝑦 ′ (0) = −11 models the motion of the body. The
discriminant of the characteristic equation is 𝐷 = 492 −4⋅4⋅49 = 1617 > 0
so the motion is overdamped free motion.
Answer: 4𝑦 ′′ (𝑡) + 49𝑦 ′(𝑡) + 49𝑦 = 0, 𝑦(0) = .5, 𝑦 ′ (0) = −11, overdamped
free motion.
11. A body weighing 16 pounds will stretch a spring 6 inches. This same body
is attached to such a spring with an accompanying dashpot. Suppose the
damping force is 4 pounds when the velocity of the body is 2 feet per
second. At 𝑡 = 0 the mass is pulled down from its equilibrium position
a distance of 1 foot and released with a downward velocity of 1 foot per
second.
Solution: The mass is 𝑚 = 16/32 = 1/2 slugs. The spring constant
𝑘 is given by 𝑘 = 16/(6/12) = 32. The damping constant is given by
𝜇 = 4/2 = 2. Since no external force is mentioned we may assume it is
zero. The initial conditions are 𝑦(0) = 1 and 𝑦 ′ (0) = 1. The following
equation 12 𝑦 ′′ (𝑡) + 2𝑦 ′ (𝑡) + 32𝑦 = 0, 𝑦(0) = 1, 𝑦 ′ (0) = 1 models the
motion of the body. The discriminant of the characteristic equation is
𝐷 = 22 − 4 ⋅ (1/2) ⋅ 32 = −60 < 0 so the motion is underdamped free
motion.
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3 Second Order Constant Coefficient Linear Differential Equations
Answer: 12 𝑦 ′′ (𝑡) + 2𝑦 ′ (𝑡) + 32𝑦 = 0, 𝑦(0) = 1, 𝑦 ′ (0) = 1, underdamped
free motion.
12. A body weighing 32 pounds will stretch a spring 2 foot. This same body
is attached to such a spring with an accompanying dashpot. Suppose the
damping constant is 8 lbs s/ft. At 𝑡 = 0 the mass is pulled down from its
equilibrium position a distance of 1 foot and released.
Solution: The mass is 𝑚 = 32/32 = 1 slug. The spring constant 𝑘 is
given by 𝑘 = 32/2 = 16. Since no external force is mentioned we may
assume it is zero. The initial conditions are 𝑦(0) = 1 and 𝑦 ′ (0) = 0.
The equation 𝑦 ′′ (𝑡) + 8𝑦 ′ (𝑡) + 16𝑦 = 0, 𝑦(0) = 1, 𝑦 ′ (0) = 0 models the
motion of the body. The discriminant of the characteristic equation is
𝐷 = 82 − 4 ⋅ 1 ⋅ 16 = 0 so the motion is critically damped free motion.
Answer: 𝑦 ′′ (𝑡) + 8𝑦 ′ (𝑡) + 16𝑦 = 0, 𝑦(0) = 1, 𝑦 ′ (0) = 0, critically damped
free motion.
13. A body weighing 2 pounds will stretch a spring 4 inches. This same body
is attached to such a spring with no accompanying dashpot. At 𝑡 = 0 the
body is pushed downward from equilibrium with a velocity of 8 inches
per second.
Solution: The mass is 𝑚 = 2/32 = 1/16 slug. The spring constant 𝑘
is given by 𝑘 = 2/(4/12) = 6 and the damping constant is 𝜇 = 0. The
initial conditions are 𝑦(0) = 0 and 𝑦 ′ (0) = 8/12 = 2/3. The equation
1 ′′
′
16 𝑦 (𝑡) + 6𝑦 = 0, 𝑦(0) = 0, 𝑦 (0) = 2/3 models the motion of the body.
The motion is undamped free or simple harmonic motion.
Answer: 𝑦 ′′ (𝑡) + 96𝑦 = 0, 𝑦(0) = 0, 𝑦 ′ (0) = 2/3, undamped free or
simple harmonic motion
14. A spring is stretched 1 m by a force of 5 N. A body of mass 2 kg is
attached to the spring with accompanying dashpot. Suppose the damping
force of the dashpot is 6 N when the velocity of the body is 1 m/s. At
𝑡 = 0 the mass is pulled down from its equilibrium position a distance of
10 centimeters and given an initial downward velocity of 10 centimeters
per second.
Solution: The mass is 𝑚 = 2. The spring constant 𝑘 is given by 𝑘 =
5/1 = 5. The damping constant is given by 𝜇 = 6/1 = 6. Since no external
force is mentioned we may assume it is zero. The initial conditions are
𝑦(0) = .1 and 𝑦 ′ (0) = .1. The equation 2𝑦 ′′ (𝑡) + 6𝑦 ′ (𝑡) + 5𝑦 = 0, 𝑦(0) =
.1, 𝑦 ′ (0) = .1 models the motion of the body. The discriminant of the
characteristic equation is 𝐷 = 62 − 4 ⋅ 2 ⋅ 5 = −4 < 0 so the motion is
underdamped free motion.
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Answer: 2𝑦 ′′ (𝑡) + 6𝑦 ′ (𝑡) + 5𝑦 = 0, 𝑦(0) = .1, 𝑦 ′ (0) = .1, underdamped
free motion.