Undamped free vibration - There are various method to analyse an undamped system. Let us take the case of mass hung by a spring. it's equivalent mathematical model is shown below:
![](/CUPS/Vibrations_files/Mathematical modelling.jpg)
based on the above mathematical model the governing equation and its solution can be found by many ways. We will solve it by
(i) Newton's second law
(ii) D'Alemberts principle, and
(iii) Energy conservation method.
Solution by Newton's 2nd law
It states that the rate of change of linear momentum is proportional to the force impressed upon it and it is mathmatically written as below
ΣF=mdvdt=>md(x˙)dt, where x is the displacement.
From the above free body diagram
mg−k(Δst+x)=mx¨ ,Where Δst=static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, kΔst=mg, thus
mx¨+kx=0,orx¨+(km)x=0,- - - - -Eq 1
now if we say let (km)=ω2n,whereωn= natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that
ωn=km−−√, frequency of vibration fn=ω2Π=12Π∗km−−√, and time period off vibration is T=1fn=2Π∗mk−−√
Solution by D. Alebert's principle
Which states that => Accelerating Force + Restoring Force = 0, hence
mg−k(Δst+x)−mx¨=0 ,Where Δst=static deflection due to the weight of the hanging mass or deflection in the absence of any vibration. Hence, kΔst=mg, thus
mx¨+kx=0,orx¨+(km)x=0,- - - - -Eq 1
now if we say let (km)=ω2n,whereωn= natural frequency of vibration of the system. Thus comparing equation 1 with the fundamental equation of SHM we can say that
ωn=km−−√, frequency of vibration fn=ω2Π=12Π∗km−−√, and time period of vibration is T=1fn=2Π∗mk−−√
Solution by Energy conservation method
It states that for any vibrating system the kinetic energy at equilibrium position will be maximum and potential energy 0, where as extreme positions potential energy will be maximum and kinetic energy 0. At any intermediate position the system will have both kinetic and potential energy. However at all position the sum of kinetic and potential energy will remain same or constant. Hence, K.E+P.E=Constant. Thus equating total energy at any potion we get
12mx˙2+k[Δst+(0+x)2]x−mgx=constant- - - - - -eqn2
differentiating eqn2 , we get
mx¨x˙+kΔst+kxx˙−mg=0, since Δst=static deflection due to the weight of the hanging mass or deflection in the absence of any vibration and kΔst=mg, we get mx¨x˙+kxx˙=0 or mx¨+kx=0. This is same fundamental equation of SHM and hence
ωn=km−−√, frequency of vibration fn=ω2Π=12Π∗km−−√, and time period of vibration is T=1fn=2Π∗mk−−√
Thus wee see that it really does not matter which method we solve the problem wee will always reach to the same values. Developing above equation for a system is known as mathematical modeling which is required for vibration analysis or dynamic analysis of the system
Governing Equation or Charecteristic Equation for spring-mass system considering the mass of the spring.
Let 'L' be the length of the spring and 'ρ' be the mass of spring per unit length.
Mass of spring, M = ρL, Consider a small element 'dy' of the spring at a distance 'y from a support. The mass of this element is 'dM'.
dM=ρdy.
Considering that the free end of the spring has moved by 'x'. The displacement at the fixed point will be zero. from similar triangles, at any point, at a distance'y', the displacement is given by 'δ' , such that:
δy=xL
hence, velocity of 'dM' will be δ˙
Therefore. δ˙=x˙yL
The kinetic energy of the system will be due to mass of the spring 'M' and the hanging mass 'm'.
Total kinetic energy is KE=KEspring+KEmass
KE=∫L0 12dMδ˙2+12mx˙2
=12mx˙2+∫L0 12ρdy(x˙yL)2, replacing the value of dM and ρ˙
=12mx˙2+12ρ(x˙L)2∫L0 y2dy
=12mx˙2+12ρ(x˙L)2L33
=12mx˙2+12ρ(x˙)2L3
=12mx˙2+16M(x˙)2, since ρL=M,The total energy in the system is say, U and U=K.E+P.E
Now P.E=12kx2. Therefore
U=12kx2+12mx˙2+16Mx˙2=>U=12kx2+12x˙2(m+M3)
Differentiatin andd using enery method we get
dUdt=0=kxx˙+(m+M3)x˙x¨, or
kx+(m+M3)x¨=0=>x¨+k(m+M3)x=0, thus we get
ωn=k(m+M3)−−−−−−√, frequency of vibration fn=ω2Π=12Π∗k(m+M3)−−−−−−√,