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

 

[rwisz]

Maximum acceleration??

The maximum acceleration of the instrument can be determined by using the following equation:

amax = Xωn2sin(ωt)

Where X is the amplitude of vibration of the instrument, ωn is the natural frequency of the system, and ωt is the angular frequency of the vibration. In this case, ωt can be calculated using the formula ωt = 2πf, where f is the frequency of the vibration (in this case, 30Hz). Substituting the given values, we get:

amax = (0.001m)(2π(30Hz))2sin(2π(30Hz))

amax = 0.041m/s2

Therefore, the maximum acceleration of the instrument is 0.041m/s2.