VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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Chapter 19.4, Problem 19.112P
To determine

The amplitude of vibration should be expected when three collar are placed on the spring.

Expert Solution & Answer
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Answer to Problem 19.112P

The amplitude of vibration of the system with three collars in out-phase motion is 5.63 mm and the amplitude of vibration of the system with three collars in in-phase motion is 22.5 mm.

Explanation of Solution

Given:

Amplitude of vibration when 1 collar is on the spring is 15 mm.

Amplitude of vibration when 2 collar is on the spring is 18 mm.

Concept used:

Write the expression for amplitude of forced vibration.

xm=δst1(ωf2ωn2) ...... (1)

Here, xm is the amplitude of forced vibration, δsr is the static displacement, ωf is the forced frequency and ωn is the natural frequency of system.

Write the expression for natural circular frequency of system having mass m.

(ωn)1=km ...... (2)

Here, (ωn)1 is the natural frequency of vibration with only one disk, m is the mass of disk and k is the stiffness of spring.

Write the expression for natural circular frequency of system having mass 2m.

(ωn)2=k2m

Substitute (ωn)1 for km in above expression.

(ωn)2=(ωn)12 ...... (3)

Here, (ωn)2 is the natural frequency of vibration with mass only two disks.

Write the expression for natural circular frequency of system having mass 3m.

(ωn)3=k3m

Substitute (ωn)1 for km in above expression.

(ωn)2=(ωn)13 ...... (4)

Here, (ωn)3 is the natural frequency of vibration with mass only three disks.

Substitute (xm)1 for xm and (ωn)1 in equation (1).

(xm)1=δst1(ωf2(ωn)12) ...... (5)

Here, (xm)1 is the amplitude of vibration with 1 disk.

Substitute (xm)2 for xm and (ωn)2 in equation (1).

(xm)2=δst1(ωf2(ωn)22)

Substitute (ωn)12 for (ωn)2 in above expression.

(xm)2=δst12(ωf2(ωn)12) ...... (6)

Here, (xm)2 is the amplitude of vibration with 2 disks

Substitute (xm)3 for xm and (ωn)3 in equation (1).

(xm)3=δst1(ωf2(ωn)32)

Substitute (ωn)13 for (ωn)2 in above expression.

(xm)3=δst13(ωf2(ωn)12) ...... (7)

Here, (xm)3 is the amplitude of vibration with 3 disks.

Calculation:

In-Phase motion:

Substitute 15 mm for (xm)1 in equation (5).

15 mm=δst1(ωf2(ωn)12) ...... (8)

Substitute 18mm for (xm)2 in equation (6).

18 mm=δst12(ωf2(ωn)12) ...... (9)

Divide equation (9) by equation (8).

18 mm15 mm=δst12(ωf2(ωn)12)δst1(ωf2(ωn)12)1.2=1(ωf2(ωn)12)12(ωf2(ωn)12)1.22.4(ωf2(ωn)12)=1(ωf2(ωn)12)

Solve the above expression for ωf2(ωn)12.

ωf2(ωn)12=17

Substitute 17 for ωf2(ωn)12 in equation (8).

15=δm117δm=12.855 mm

Substitute 12.855 mm for δm and 17 for ωf2(ωn)12 in equation (7).

(xm)3=12.85513(17)=22.5 mm

The amplitude of vibration of the system with three collars in in-phase motion is 22.5 mm.

Out-of phase motion:

Substitute 18mm for (xm)2 in equation (6).

18 mm=δst12(ωf2(ωn)12) ...... (10)

Divide Equation (10) by equation (8).

18 mm15 mm=δst12(ωf2(ωn)12)δst1(ωf2(ωn)12)1.2=1(ωf2(ωn)12)12(ωf2(ωn)12)1.2+2.4(ωf2(ωn)12)=1(ωf2(ωn)12)

Solve the above expression for ωf2(ωn)12.

ωf2(ωn)12=0.647

Substitute 0.647 for ωf2(ωn)12 in equation (8).

15=δm10.647δm=5.295 mm

Substitute 5.295 mm for δm and 0.647 for ωf2(ωn)12 in equation (7).

(xm)3=5.29513(0.647)=5.63 mm

The amplitude of vibration of the system with three collars in out-phase motion is 5.63 mm.

Conclusion:

Thus, the amplitude of vibration of the system with three collars in out-phase motion is 5.63 mm and the amplitude of vibration of the system with three collars in in-phase motion is 22.5 mm.

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Chapter 19 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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