A single degree of freedom system is subjected to a harmonic
exitation.The equation of motion is   m(u")+c(u')+ku=Pcos(omega*t)
where m=2kg,c=4 N/(m-s),k=8 N/m,P=12N,omega=0.25Hz
Find:(a)the undamped natural frequency,(b)the critical damping
constant
     (c)the damping ratio (d)the damped natural frequency 
     (e)the amplitude of the steady state vibration response

Hello again wjs-ga,

I will refer again to the University of Saskatchewan, College of
Engineering website
http://www.engr.usask.ca/classes/CE/804/notes/fv_sdof.pdf


a) Undamped natural frequency, Wo = SQRT(K/M)  (See equation 2.5)

In your problem, K = 8 N/m; M = 2 kg

Note: 1 N = 1 kg–m/sec^2   ==>  1 N/m = 1 kg/sec^2

==> Wo = SQRT[(8 kg/sec^2)/2 kg] = SQRT[4(1/sec^2)] = 2 rad/sec



b) Critical damping constant, Ccrit = 2MWo    (equation 2.32b)

==> Ccrit = 2(2 kg)(2 rad/sec) = 8 kg/sec = 8 N–Sec/m


c) Damping ratio, Psi = C/Ccrit    (equation 2.44)

In your problem, C = 4 N–Sec/m

==> Psi = 4/8 = 0.5 (dimensionless)



d) Damped natural frequency, Wd = Wo(SQRT(1-Psi^2))     (equation
2.46)

This gives you Wd = (2 rad/sec)(SQRT(1-(.5)^2)) = 1.732 rad/sec



e) Amplitude of the steady state vibration response.

For this one, refer to another page at the University of Saskatchewan,
College of Engineering website
http://www.engr.usask.ca/classes/CE/804/notes/har_sdof_ovh.pdf


Based on equation 3.2, the amplitude is 

                                       1
(Fo/K)(H(w)) where H(w) = -------------------------------------
                           SQRT[((1-(w/Wo)^2)^2+[2Psi(w/Wo)]^2]


For your problem, this gives you

                    1
(12/8)[-----------------------------------------] = 1.512 m
       SQRT[((1-(.25/2)^2)^2+[2(0.5)(0.25/2)]^2]



Other references:

Kettering University
http://www.gmi.edu/~drussell/Demos/SHO/mass-force.html


Faculty of Technology, University of Plymouth
http://www.tech.plym.ac.uk/sme/mech226/forcedamp2.pdf


Mechanical Vibrations, Second Edition, Singiresu S. Rao
Addison-Wesley Publishing Company, 1990


I hope you have found this information helpful. If you have any
questions, please request clarification prior to rating the answer.

Googlenut


Google Search Terms:

damped forced vibration natural frequency
://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&safe=off&q=damped+forced+vibration+natural+frequency&btnG=Google+Search

damped forced vibration
://www.google.com/search?hl=en&ie=ISO-8859-1&q=damped+forced+vibration&btnG=Google+Search

---

Request for Answer Clarification by wjs-ga on 10 Mar 2003 23:03 PST

what's the w in  H(w) = ------------------------------------- 
                           SQRT[((1-(w/Wo)^2)^2+[2Psi(w/Wo)]^2]

---

Clarification of Answer by googlenut-ga on 10 Mar 2003 23:22 PST

H(w) means H is a function of frequency, w. I'm sorry it wasn't clear.
 I probably wasn't consistent with my terminology throughout the
problem.

In this case, w is what you called omega, which is 0.25.

The University of Plymouth reference has another form of that equation
that might be easier to follow.

It's getting late here. If you have any more questions, I may not be
able to get to them until the morning.  I hope that doesn't cause you
any inconvenience.

Googlenut