The steady state amplitude of motion of the mass.

Question

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Answer 1

Question

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

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Answer 2

Question

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

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Answer 3

Question

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

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Answer 4

Question

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

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Answer 5

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

Answer 6

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Expert Solution & Answer

Answer to Problem 13.1P

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

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

Answer 7

Chapter 13, Problem 13.1P

To determine

The steady state amplitude of motion of the mass.

Answer 8

Expert Solution & Answer

Answer to Problem 13.1P

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

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

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

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,

$y(t)=Ysinωtω=2πff=3 Hzω=6π rad/sec$

Substitute the value in the equation,

We get,

$y(t)=4×10−3sin6πt$

Natural frequency is given as

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

Frequency ratio is given as:

$r=ωωn$

Squaring both sides, we get,

$r2=6π21000r2=0.3553$

Displacement transmissibility for base excitation is given as:

$XY=4ζ2r2+1(1−r2)2+4ζ2r2$

Steady state amplitude of motion of $X$ is given as:

$X=Y4ζ2r2+1(1−r2)2+4ζ2r2$

Neglect the damping, $ζ=0$

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

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

Substitute $r2=0.3553$ and $Y=4×10−3$

$X=4×10−31(1−0.3553)X=6.20×10−3 mX=6.20 mm$

Conclusion:

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

Answer 12

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The steady state amplitude of motion of the mass.

Concept explainers

The steady state amplitude of motion of the mass.

Answer to Problem 13.1P

Explanation of Solution

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