How would g(gravitaional acceleration) affect vibration? 1

[kyong]

This is not a practical problem. Rather, it is a problem to understand vibration in general sense a little more.

To calculate fundamental period of vibration, I often use Rayleigh approximation which is

T = 2* pi * [(summation(W*y^2)/(g*summation(W*y))]^0.5

where W=weight of each segment, y=deflection of each segment from stattic equilibrium position


Apparently it seems that g is a factor such that on the moon T may be much different. But if direction of vibtation is horizontal, I'm sure T will be the same. If direction of vibration is vertical, what will happen? Please tell me something I can learn. Thanks.

kyong

 

[magicme]

hello kyong

in the sense that you are talking about, gravitational acceleration does not effect natural frequencies, as the basic formula is freq = sqrt( stiffness / Mass ) with frequency in radians / sec.

if you moved your system to the moon, yes g would change but so would the weight W by a proportional amount....it is the mass thats important and that doesnt change.

in a more complex train of thought.... acceleration CAN effect natural frequencies. for example, the natural frequencies of a propellor or turbine blade increase as rotational speed increases because there is a "stiffening" radial force due to rotation that tends to inhibit lateral vibrations....so the natural frequencies of turbine blades and propellors and such increase as rotational speed increases.

daveleo

 

[tomirvine]

There are also some special cases the natural frequency depends on gravity, such as a pendulum or inverted pendulum.

Tom Irvine

http://www.vibrationdata.com

 

[IRstuff]

Auto suspensions are designed so that the natural frequency is a direct result of the compression of the suspension by the sprung mass.

TTFN

 

[GregLocock]

If I can have a go

the compression of the springs in the system due to gravity is y=m*g/k (1)


m=W/g

Rayleigh says 1/2*m*r^2*w^2=1/2*k*r^2

so 1/w=sqrt(m/k)

w=2*pi/T

so

T=2*pi*sqrt(m/k)

So substituitng for k

T=2*pi*sqrt(m*y/m/g)



T=2*pi*sqrt(y/g)

BUT the only reason that g is in there is because y is proportional to g, that is, if you reduce g to 1.6 ms-2, then y would drop by the same factor, and the period of oscillation would remain the same.





Cheers

Greg Locock

 

[kyong]

Thank you, guys.
Your inputs are definitely helpful to me.

kyong