An underdamped shock absorber is to be designed for a motor cycle of mass 200 kg(m). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated below. Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 10 sec(τd) and the amplitude x1is to be reduced to one-tenth in one half cycle.
An underdamped shock absorber is to be designed for a motor cycle of mass 200 kg(m). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated below. Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 10 sec(τd) and the amplitude x1is to be reduced to one-tenth in one half cycle.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
An underdamped shock absorber is to be designed for a motor cycle of mass 200 kg(m). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated below. Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 10 sec(τd) and the amplitude x1is to be reduced to one-tenth in one half cycle. Example is attached
![X1.5
td
=
@₁ =
n
Example 2-14: Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a motor cycle of
mass 200 kg (m). When the shock absorber is subjected to an initial vertical
velocity due to a road bump, the resulting displacement-time curve is to be as
indicated below. Find the necessary stiffness and damping constants of
the shock absorber if the damped period of vibration is to be 2 sec (T)
and the amplitude x₁ is to be reduced to one-fourth in one half cycle
X₁
(i.e., X₁.5 ).
4
4
2
2π
@d
X2
=
k/2
700000,
c = 2√√km =
Cc
X1.5X₁
=
4
16
2
m
mm
k/2
x(t)
O
... →8 = ln
→@j=7, @= @√1-5² -
d
X1.5
X₁
X₂
→
=
= ln(16) = 2.7726 =
= π → @n
x2.5
= 2√√(2,358.2652)(200) = 1,373.54
=
k
2
→ k = mw² = (200)(3.4338)² = 2,358.2652
n
m
с
2 = · → c = c = (0.4037)(1,373.54)=554.4981
T
1-(0.4037)²
N
N• sec
m
m
S 8 =
2πζ
√₁-5²
2πζ
2
√1-5²
N • sec
→=0.4037
= 3.4338
rad
sec](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20000a63-5a8a-4b5e-adbd-85ecbf6c298b%2F76cf2bbf-87c1-425c-8696-75d3b72cb9a7%2Fk38iitc_processed.png&w=3840&q=75)
Transcribed Image Text:X1.5
td
=
@₁ =
n
Example 2-14: Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a motor cycle of
mass 200 kg (m). When the shock absorber is subjected to an initial vertical
velocity due to a road bump, the resulting displacement-time curve is to be as
indicated below. Find the necessary stiffness and damping constants of
the shock absorber if the damped period of vibration is to be 2 sec (T)
and the amplitude x₁ is to be reduced to one-fourth in one half cycle
X₁
(i.e., X₁.5 ).
4
4
2
2π
@d
X2
=
k/2
700000,
c = 2√√km =
Cc
X1.5X₁
=
4
16
2
m
mm
k/2
x(t)
O
... →8 = ln
→@j=7, @= @√1-5² -
d
X1.5
X₁
X₂
→
=
= ln(16) = 2.7726 =
= π → @n
x2.5
= 2√√(2,358.2652)(200) = 1,373.54
=
k
2
→ k = mw² = (200)(3.4338)² = 2,358.2652
n
m
с
2 = · → c = c = (0.4037)(1,373.54)=554.4981
T
1-(0.4037)²
N
N• sec
m
m
S 8 =
2πζ
√₁-5²
2πζ
2
√1-5²
N • sec
→=0.4037
= 3.4338
rad
sec
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