Engineering Mathematics-I

Lecture 39 : Second Order ODE: Applications

Panchatcharam Mariappan1

1 Assistant Professor

Department of Mathematics and Statistics

IIT Tirupati, Tirupati
Overdamping, Critical
damping,
Underdamping

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Second-Order ODE: Simple Harmonic

Figure 1: Source: Kreyzig Book

Example 1
Assume an object weighing 9.8 kg (scientifcailly Newton) stretches a spring 20
cm. Find the equation of motion if the spring is released from the equilibrium
position with an upward velocity of 14 m/sec. What is the period of the motion? 2
Second-Order ODE: Simple Harmonic
Modelling Part:
• Hooke’s law F1 = −ky
• Newton’s second law F = my ′′
• F = F1 =⇒ my ′′ + ky = 0
• W = mg = 9.8 stretches s = 20/100 = 1/5
• mg = ks =⇒ k = 49
• W = mg =⇒ 9.8 = m(9.8) =⇒ m = 1
• y ′′ + 49y = 0
• y(0) = 0, y ′ (0) = −14

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Undamped System

• Natural Frequency f = 2πω

of the system
√
• Alternative way y = C cos(ωt − θ), C = A2 + B 2 , θ = tan−1 B


A 4
Second-Order ODE: Simple Harmonic
Solution Part:
• Time dependent problem, so y is a function of t
q
• Auxiliary equation r2 + mk
= 0 =⇒ r = ±iω, ω = k
m
• y(t) = A cos ωt + B sin ωt
• In our problem, k = 49, m = 1 =⇒ ω = 7
• y(t) = A cos 7t + B sin 7t
• y(0) = 0 =⇒ A = 0
• y ′ (0) = −14 =⇒ B = −2
• y(t) = −2 sin 7t

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Second-Order ODE: Simple Harmonic
Solution Part:

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Second-Order ODE: Motorbike Suspension

Figure 2: Source: Kreyzig Book and

https://www.isotechinc.com/product-category/air-cylinders-actuators/dashpots/

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Second-Order ODE: Lander Suspension

Figure 3: Source: https://en.wikipedia.org/wiki/Phoenix_(spacecraft)

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Second-Order ODE: Motorbike Suspension
• Real world problem has always some friction in the system.
• Friction - Oscillations die slowly - Damping
• Air resistance, physical damper or dashpot
• Damping is a friction force, proportional to the velocity of the mass and
acts in the opposite direction
• cy ′ be the damping force
• my ′′ + cy ′ + ky = 0 =⇒ y ′′ + c ′
my + k
my =0

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Second-Order ODE: Motorbike Suspension
• Auxiliary equation: r2 + c
mr + k
m =0
• r1 = −α + β, r2 = −α − β
√
• c
α = 2m 1
, β = 2m c2 − 4mk
• Overdamping: c2 > 4mk
• Critical Damping: c2 = 4mk
• Underdamping: c2 < 4mk

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Overdamping: c2 > 4mk

• y(t) = c1 e−(α−β)t + c2 e−(α+β)t

• Both exponents are negative. For, α > 0, β > 0, β 2 < α2
• As t → ∞, y(t) → 0. Static equilibrium position 11
Critically Damped System: c2 = 4mk

• Similar to Overdamping, does not oscillate, however, if damping is

reduced a little, it oscillates.
• Physical systems are either over or underdamped
• β = 0, r1 = r2 = −α
• y(y) = (c1 + c2 t)e−αt
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Under Damped System: c2 < 4mk

• Roots are complex. y1 = −α + iω, y2 = −α − iω

q
• ω= m k c2
− 4m 2

• y(t) = e (A cos ωt + B cos ωt) = Ce−αt A cos(ωt − θ)

−αt

• Curve lies between Ce−αt and −Ce−αt 13

Second-Order ODE: Motorbike Suspension

Figure 4: Souce: https://www.heromotocorp.com/en-in/the-bike/super-splendor-bs6-101.html

Example 2
A herohonda bike weighs 109 kg (scientifally Newton), and we assume a rider
weight of 67.4 kg. When the rider mounts the bike, the suspension compresses
9.8 cm., then comes to rest at equilibrium. The suspension system provides
damping equal to 360 times the instantaneous vertical velocity of the bike (and
rider). 14
Second-Order ODE: Motorbike Suspension
Modelling Part:
• W = mg = 109 + 67.4 = 176.4 stretches s = 9.8/100 = 0.098
• mg = ks =⇒ 176.4 = 0.098k =⇒ k = 1800
• W = mg =⇒ 176.4 = m(9.8) =⇒ m = 18
• 18y ′′ + 360y ′ + 1800y = 0
• When the bike was in the air before contacting the ground, the wheel was
hanging freely and the spring was uncompressed. Therefore, the wheel is
0.098m below the equilibrium position
• y(0) = 0.098 =⇒ 0.098 = c1

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Second-Order ODE: Motorbike Suspension
Solution Part: Let us solve the problem and find the period of motion once
we know how to solve the equation.
• c = 360, m = 18, k = 1800 =⇒ c2 = 129600 = 4mk.
• If the bike hits the ground with velocity, 10m/sec downward, then
y ′ (0) = 10
• y ′ = c2 e−10t − 10(c1 + c2 t)e−10t
• 10 = c2 − 10c1 =⇒ c2 10.98
• y(t) = (0.098 + 10.98t)e−10t

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Second-Order ODE: Motorbike Suspension
Solution Part: Let us solve the problem and find the period of motion once
we know how to solve the equation.

• y(t) = (0.098 + 10.98t)e−10t

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RLC Circuit

• A current I in the circuit causes a voltage drop RI across the resistor

(Ohm’s law) and a voltage drop LI ′ across the conductor,
• sum of these two voltage drops equals the EMF (Kirchhoff’s Voltage Law)

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RLC Circuit

R E
I′ + I=
L L
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RLC Circuit

• An RLC-circuit is obtained from an RL-circuit by adding a capacitor

• the voltage drop Q/C across the capacitor
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RLC Circuit

Q
LI ′ + RI +
= E(t)
C 
′
I = Q =⇒ Q = I
I
LI ′′ + RI ′ + = E ′ (t)
C
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RLC Circuit
E ′ (t) = E0 ω cos ωt

A(D) = (D2 + ω 2 )
=⇒ Ip = A cos ωt + B sin ωt
=⇒ Ip′ = ω(−A sin ωt + B cos ωt)
=⇒ Ip′′ = −ω 2 (A cos ωt + B sin ωt)
Ip
=⇒ LIp′′ + RIp′ + = E0 ω cos ωt
C
A
=⇒ Lω 2 (−A) + RωB + = E0 ω
C
B
and Lω 2 (−B) + Rω(−A) + =0
C

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RLC Circuit

1
Reactance S = ωL −
ωC
−SA + RB = E0
−RA − SB = 0
−E0 S
A= 2
R + S2
E0 R
B= 2
R + S2
p
Impedance = R2 + S 2

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NASA Lander System(Exercise)

NASA is planning a mission to Mars. To save money, engineers have decided

to adapt one of the moon landing vehicles for the new mission. However, they
are concerned about how the different gravitational forces will affect the
suspension system that cushions the craft when it touches down. The
acceleration resulting from gravity on the moon is 1.6m/sec2 , whereas on
Mars it is 3.7m/sec2 .
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NASA Lander System
• The lander has a mass of 15,000 kg and the spring is 2 m long when
uncompressed. The lander is designed to compress the spring 0.5 m to
reach the equilibrium position under lunar gravity. The dashpot imparts a
damping force equal to 48,000 times the instantaneous velocity of the
lander. Set up the differential equation that models the motion of the
lander when the craft lands on the moon.
• Let time t = 0 denote the instant the lander touches down. The rate of
descent of the lander can be controlled by the crew, so that it is
descending at a rate of 2 m/sec when it touches down. Find the equation
of motion of the lander on the moon.

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NASA Lander System
• If the lander is traveling too fast when it touches down, it could fully
compress the spring and “bottom out.” Bottoming out could damage the
landing craft and must be avoided at all costs. Graph the equation of
motion found in part 2. If the spring is 0.5 m long when fully compressed,
will the lander be in danger of bottoming out?
• Assuming NASA engineers make no adjustments to the spring or the
damper, how far does the lander compress the spring to reach the
equilibrium position under Martian gravity?

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NASA Lander System
• If the lander crew uses the same procedures on Mars as on the moon,
and keeps the rate of descent to 2 m/sec, will the lander bottom out
when it lands on Mars?
• What adjustments, if any, should the NASA engineers make to use the
lander safely on Mars?

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Power Series

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Power Series
When the linear ODE contains variable coefficients, then power series method
is very helpful to solve the problem. The solution of the ODE is in the form of
power series.

Definition 1 (Power Series)

A power series (in powers of (x − x0 )) is an infinite series of the form
∞
X
am (x − x0 )m = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + · · · (1)
m=0

Here x is a variable. a0 , a1 , a2 , · · · are constants, called the coefficient of the

series. x0 is a constant, called the center of the series.

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 Thanks
Doubts and Suggestions
panch.m@iittp.ac.in

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Engineering Mathematics-I
Lecture 39 : Second Order ODE: Applications

Panchatcharam Mariappan1

1 Assistant Professor

Department of Mathematics and Statistics

IIT Tirupati, Tirupati

30