System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 13, Problem 13.1P
To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer
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Answer to Problem 13.1P

The steady state amplitude of motion of the mass is X=6.20mm.

Explanation of Solution

Given:

m=0.5kgk=500N/mY=4mmf=3Hz

System Dynamics, Chapter 13, Problem 13.1P

Concept Used:

Displacement transmissibility.

Calculation:

The equation of motion of mass is given as:

mx+kx=ky

Where,

y(t)=Ysinωtω=2πff=3Hzω=6πrad/sec

Substitute the value in the equation,

We get,

y(t)=4×103sin6πt

Natural frequency is given as

ωn=kmωn2=kmωn2=5000.5ωn=1000rad/sec=31.62rad/sec

Frequency ratio is given as:

r=ωωn

Squaring both sides, we get,

r2=6π21000r2=0.3553

Displacement transmissibility for base excitation is given as:

XY=4ζ2r2+1(1r2)2+4ζ2r2

Steady state amplitude of motion of X is given as:

X=Y4ζ2r2+1(1r2)2+4ζ2r2

Neglect the damping, ζ=0

Replace the value of ζ=0 in the above equation, we get

X=Y0+1(1r2)2+0X=Y1(1r2)2X=Y1(1r2)

Substitute r2=0.3553 and Y=4×103

X=4×1031(10.3553)X=6.20×103mX=6.20mm

Conclusion:

The steady state amplitude of motion of the mass is X=6.20mm.

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