# Spring Mass Damping System Question? Maximum acceleration?

Spring Mass Damping System Question?? Maximum acceleration??

## Homework Statement

Hello,

I was wondering if anyone knows how to go about answering these type of questions...

Anti-vibration mounts are used to attach an instrument of mass 5kg to a panel. The panel is vibrating with an amplitude of 1mm at a frequency 30Hz.
Determine
a)the stiffness of the mounts which provides an isolation effect, i.e a reduction in the vibration amplitude of the attached instrument...

b)the maximum acceleration to which the instrument is exposed when the mounts have an effective stiffness of 30kN/m and also provide viscous damping with a damping coefficient of 60Ns/m

c)the acceleration amplitude of the instrument at resonance

## Homework Equations

Given MF= X/Y = √( (1 +(2zr)2)/((1-r2) 2 + (2zr)2) )

where X- instrument amplitude of vibration
Y- panel amplitude of vibration
r- frequency ratio
z- damping ratio
MF- magnification factor

We have the following equations

Critical Damping Coefficient cc = 2√km = 2mωn

Equations

f = ωn / 2*(pi)

ωn = √k/m

cc = 2√km

z = c/cc

ωd = ωn√(1-z2)

## The Attempt at a Solution

--------------------

Solutions

a) Post 3 for solution

b) Post 4 for solution

c)??

Last edited:

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
rock.freak667
Homework Helper

Your method for the first one is correct.

For the second part I think you can just use amax = ω2Xmax.

Here is the solution for part a, its just that we have to take into consideration that r > √2 otherwise no isolation

a) For isolation to happen r > √2

Now r = ω/ωn

Now forced frequency ω = 2(pi)f = 2*pi*30 = 188.49 rad/s

so for r > √2

ω/ωn > √2
ω/√2 > ωn as ω =188.49

188.49/√2 > ωn

133.28 > ωn now if ωn = √(k/m)

133.28 > √(k/m) where m =5 kg

133.28 > √(k/5)

133.282 > (k/5)

17764.2 > (k/5)

5*17764.2 > k

88826>k for isolation

88821 N/m >k

also thanks rockfreak your right about that,

any ideas about part c...

Now the solution for b)

We have the following equations

Critical Damping Coefficient cc = 2√km = 2mωn

f = ωn / 2*(pi)

ωn = √k/m

cc = 2√km

z = c/cc

ωd = ωn√(1-z2)

Therefore we can get

ωn = √k/m = √30*103/5 = 77.45 rad/s

And from above equations we can derive that

z = c/cc = c/2mωn

We are given damping coefficient of 60 Ns/m

therefore z = 60/2*5*77.45 = 0.077459

and r = ω/ωn = 188.49/77.45 = 2.43

Now putting these values in the magnification factor equation we can derive the amplitude X of the instrument therefore

Given MF= X/Y = √( (1 +(2zr)2)/((1-r2)2 + (2zr)2) )

So

X/Y = √( (1 +(2*0.077459*2.43)2)/((1-2.432)2 + (2*0.077459*2.43)2) )

X/Y = √(0.04717 ) now Y = 0.001

therefore X = 0.0002172m

Now if maximum acceleration

amax = Xω2 = 0.0002172 * 188.492 = 7.71 m/s2

Last edited: