What Is the Minimum Damping Constant to Limit Vibration Amplitude to 3.2 mm?

[Firben]

Homework Statement

A sensitive instrument with the mass of 30 kg should be installed in a labratory. In the labratory, there exist a machine that causes the floor to vibrate with the frequency of 0.8 Hz and with the amplitude 8.0 mm, each with the spring constant k = 500 N/m. Its then connected to a damper according to the figure.

http://s716.photobucket.com/user/Pitoraq/media/Mek22_zpsec302081.png.html
(note the 4 springs)Determine the smallest value that the damping constant c can have so that the vibration amplitude for the instrument due to the floors vibration never go over 3.2 mm

Homework Equations

X = b(ω/ω_n)^2/√[1-(ω/ω_n)^2)^2+[2ζω/ω_n]^2) (1)
ω_n = √(4k/m) (2)
ζ = c/(2mω_n) (3)

ω = 2πf = 5.026 rad/s
ω_n = 8.165 rad/s

X = 8.0 mm
b = 3.2 mm

The Attempt at a Solution

I started with (1) to solve for the damping factor ζ, but when i did that and plugged the values in i only got imaginary numbers. I am not sure if i done it right. Since i only need to know ζ to solve for c in (3)

 

[rude man]

0.8Hz is not the natural frequency. At the natural frequency, without damping, the amplitude would be infinite!

The frequency is 0.8Hz with and without damping. It's the frequency of the forcing function in your diff. equations.

Determine the forcing function necessary to produce an amplitude of 8.0 mm without the damper, then use that forcing function to determine c needed to reduce the amplitude from 8.0 mm to 3.2 mm.

P.S. I see only 2 springs. Are you sure there are four?

 

[SteamKing]

0.8Hz is not the natural frequency. At the natural frequency, without damping, the amplitude would be infinite!

It doesn't look like the OP assumed 0.8 Hz as the natural frequency. According to his calculations, the natural frequency is 8.165 rad/s, which is equivalent to 1.3 Hz, as obtained from plugging the values m = 30 kg and k = 500 N/m into Eq. (2).

The frequency is 0.8Hz with and without damping. It's the frequency of the forcing function in your diff. equations.

Determine the forcing function necessary to produce an amplitude of 8.0 mm without the damper, then use that forcing function to determine c needed to reduce the amplitude from 8.0 mm to 3.2 mm.

P.S. I see only 2 springs. Are you sure there are four?

Looking at the diagram, you see only one view of the test apparatus. I think you can assume the test device has a spring at each corner.

 

[rude man]

It doesn't look like the OP assumed 0.8 Hz as the natural frequency. According to his calculations, the natural frequency is 8.165 rad/s, which is equivalent to 1.3 Hz, as obtained from plugging the values m = 30 kg and k = 500 N/m into Eq. (2).

I agree. I should have been more careful ...

 

[Firben]

Does anyone know how to solve this ?

What should i do ?

Should i solve for wn ?

 

[rude man]

The big question is whether your equations are correct. Either they're not or you made a math mistake somewhere.

The basic approach is to write the differential equation for the system: ∑ F = ma = mx''. Your initial conditions on x and x' are zero. You have a forcing function on x = x_f = 8mm sinωt where ω = 2π(0.8 Hz).

I.e. you have something like mx'' + kx + cx' = m d²/dt²{x_f}.

k is the EFFECTIVE spring constant. With 4 springs what would that be?

You then solve for x(t) and then solve that equation for c given x(t) = 3.2 mm max.

If you're not sufficiently familiar with ordinary linear constant-coefficient differential equations you'll have to either go with what you were taught or use the Web where there are lots of good analyses of this sort of problem. NOTE: you are interested in the STEADY-STATE solution only.

E.g http://hyperphysics.phy-astr.gsu.edu/hbase/oscdr.html