Spring Mass Damping System Question? Maximum acceleration?

[Someone121]

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 c_c = 2√km = 2mω_n

Equations

f = ω_n / 2*(pi)

ω_n = √k/m

c_c = 2√km

z = c/c_c

ω_d = ω_n√(1-z²)

The Attempt at a Solution

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

Solutions

a) Post 3 for solution

b) Post 4 for solution

c)??

 

[rock.freak667]

Your method for the first one is correct.

For the second part I think you can just use a_max = ω²X_max.

 

[Someone121]

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/ω_n

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

so for r > √2

ω/ω_n > √2
ω/√2 > ω_n as ω =188.49

188.49/√2 > ω_n

133.28 > ω_n now if ω_n = √(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...

 

[Someone121]

Now the solution for b)

We have the following equations

Critical Damping Coefficient c_c = 2√km = 2mω_n

f = ω_n / 2*(pi)

ω_n = √k/m

c_c = 2√km

z = c/c_c

ω_d = ω_n√(1-z²)

Therefore we can get

ω_n = √k/m = √30*10³/5 = 77.45 rad/s

And from above equations we can derive that

z = c/c_c = c/2mω_n

We are given damping coefficient of 60 Ns/m

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

and r = ω/ω_n = 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

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